JEE (Main) · 2024
JEE (Main) 2024 — 8 April, Shift 2
Which chapter each question came from, and how to solve each one. Keep the official paper open beside this page.
01
Summary
The heaviest chapters in each subject, and how the questions were spread by level.
Mathematics
30 Q
- Integral Calculus3
- Straight Lines3
- Sets, Relations and Functions2
8 easy · 22 medium · 0 hard
Physics
30 Q
- Kinematics2
- Laws of Motion2
- Properties of Bulk Matter2
18 easy · 11 medium · 1 hard
Chemistry
30 Q
- Organic Chemistry: Some Basic Principles and Techniques5
- Chemical Bonding and Molecular Structure2
- Electrochemistry2
21 easy · 9 medium · 0 hard
02
Questions
Tap Explain on any question for a short summary and our full working.
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | Sets, Relations and FunctionsEquivalence relations | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctMedium | |
A medium single correct question on equivalence relations, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Work modulo 2: $3\equiv 1$ and $7\equiv 1$, so $3ad-7bc$ is even exactly when $ad-bc$ is even, i.e. when $ad\equiv bc \pmod 2$.
Reflexive: $(a,b)R(a,b)$ needs $3ab-7ab=-4ab$ even, which is always true.
Symmetric: $(c,d)R(a,b)$ needs $cb\equiv da \pmod 2$, the same condition as $ad\equiv bc$. So R is symmetric.
Transitive: take $(3,1)$, $(2,4)$, $(2,1)$.
$(3,1)R(2,4)$: $3\cdot3\cdot4-7\cdot1\cdot2=36-14=22$, even.
$(2,4)R(2,1)$: $3\cdot2\cdot1-7\cdot4\cdot2=6-56=-50$, even.
$(3,1)R(2,1)$: $3\cdot3\cdot1-7\cdot1\cdot2=9-14=-5$, odd. So R is not transitive.
R is reflexive and symmetric but not transitive.
Answer: (D). | ||||
| 2 | Sets, Relations and FunctionsOne-one, into and onto functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctMedium | |
A medium single correct question on one-one, into and onto functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Find $g$ piece by piece.
$x=0$: $f(0)=-a$, $|f(0)|=a$, so $g(0)=\frac{-a-a}{2}=-a$.
$-a\le x<0$: $|x|=-x\in(0,a]$, so $f(|x|)=-x+a$; $f(x)=-a$, $|f(x)|=a$. Then $g(x)=\frac{-x+a-a}{2}=-\frac{x}{2}$, which takes values in $(0,\frac{a}{2}]$.
$0<x\le a$: $f(|x|)=f(x)=x+a>0$, so $g(x)=\frac{(x+a)-(x+a)}{2}=0$.
Not one-one: $g$ is $0$ at every point of $(0,a]$.
Not onto: the range is $\{-a\}\cup(0,\frac{a}{2}]\cup\{0\}$, which misses, for example, $-\frac{a}{2}$ and $a$.
So $g$ is neither one-one nor onto.
Answer: (D). | ||||
| 3 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | Single correctEasy | |
A easy single correct question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants For infinitely many solutions the coefficient determinant must vanish:
$\Delta=\begin{vmatrix}1&4&-1\\7&9&\mu\\5&1&2\end{vmatrix}=1(18-\mu)-4(14-5\mu)-1(7-45)=19\mu=0$, so $\mu=0$.
With $\mu=0$ the left side of the second equation is a combination of the other two: $2(x+4y-z)+(5x+y+2z)=7x+9y$.
For the system to be consistent the right sides must combine the same way: $2\lambda+(-1)=-3$, so $\lambda=-1$.
Then the three planes share a line, giving infinitely many solutions.
$2\mu+3\lambda=0-3=-3$.
Answer: (B). | ||||
| 4 | Matrices and DeterminantsDeterminants of order two and three | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on determinants of order two and three, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Subtract the first row from the second and third rows:
$\begin{vmatrix}\alpha&b&c\\a-\alpha&\beta-b&0\\a-\alpha&0&\gamma-c\end{vmatrix}=0.$
Expanding along the first row:
$\alpha(\beta-b)(\gamma-c)-b(a-\alpha)(\gamma-c)-c(a-\alpha)(\beta-b)=0$, i.e.
$\alpha(\beta-b)(\gamma-c)+b(\alpha-a)(\gamma-c)+c(\alpha-a)(\beta-b)=0.$
Divide by $(\alpha-a)(\beta-b)(\gamma-c)\ne0$:
$\frac{\alpha}{\alpha-a}+\frac{b}{\beta-b}+\frac{c}{\gamma-c}=0.$
Since $\frac{\alpha}{\alpha-a}=1+\frac{a}{\alpha-a}$: $\frac{a}{\alpha-a}+\frac{b}{\beta-b}+\frac{c}{\gamma-c}=-1$.
Now $\frac{\gamma}{\gamma-c}=1+\frac{c}{\gamma-c}$, so
$\frac{a}{\alpha-a}+\frac{b}{\beta-b}+\frac{\gamma}{\gamma-c}=-1+1=0.$
Answer: (B). | ||||
| 5 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | Single correctMedium | |
A medium single correct question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations MATHEMATICS has 11 letters: M, A, T twice each, and H, E, I, C, S once each, so 8 distinct letters of which 3 can be taken twice.
Count selections of 5 letters:
All 5 different: $\binom{8}{5}=56$.
One letter twice and 3 others different: $\binom{3}{1}\binom{7}{3}=3\times35=105$.
Two letters twice and 1 other: $\binom{3}{2}\binom{6}{1}=3\times6=18$.
(No letter can appear three times.)
Total $=56+105+18=179$.
Answer: (C). | ||||
| 6 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | Single correctEasy | |
A easy single correct question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem General term: $T_{r+1}=\binom{10}{r}(\sqrt{a}x^2)^{10-r}\left(\frac{1}{2x^3}\right)^r=\binom{10}{r}\frac{a^{(10-r)/2}}{2^r}x^{20-5r}$.
Independent of $x$: $20-5r=0$, so $r=4$.
$T_5=\binom{10}{4}\frac{a^{3}}{2^4}=\frac{210a^3}{16}=105$, so $a^3=8$ and $a=2$.
$a^2=4$.
Answer: (B). | ||||
| 7 | Complex NumbersAlgebra of complex numbers | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on algebra of complex numbers, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations Let $c=\cos\theta$. Multiply numerator and denominator by $1+2ic$:
$\frac{(1+ic)(1+2ic)}{1+4c^2}=\frac{1-2c^2+3ic}{1+4c^2}.$
Purely imaginary: real part $1-2c^2=0$, so $\cos^2\theta=\frac12$, $\cos\theta=\pm\frac{1}{\sqrt2}$ (then the imaginary part $3c\ne0$).
In $[-\pi,2\pi]$: $\theta=-\frac{3\pi}{4},-\frac{\pi}{4},\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}$.
Sum $=\left(-\frac{3\pi}{4}-\frac{\pi}{4}+\frac{\pi}{4}+\frac{3\pi}{4}\right)+\frac{5\pi}{4}+\frac{7\pi}{4}=0+3\pi=3\pi$.
Answer: (B). | ||||
| 8 | Sequences and SeriesGeometric progressions | Class 11: Sequences and Series | Single correctEasy | |
A easy single correct question on geometric progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series Let the terms be $ar^{n-1}$ with $a>0$, $r>1$.
$T_3T_5=a^2r^6=49$, so $ar^3=7$ (the fourth term).
$T_2+T_6=ar+ar^5=\frac{7}{r^2}+7r^2=\frac{70}{3}$, so $r^2+\frac{1}{r^2}=\frac{10}{3}$, giving $r^2=3$ or $\frac13$. Increasing, so $r^2=3$.
$T_4+T_6+T_8=ar^3(1+r^2+r^4)=7(1+3+9)=91$.
Answer: (C). | ||||
| 9 | Limits, Continuity and DifferentiabilityContinuity | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on continuity, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Left limit: $\lim_{x\to0^-}\frac{\tan((a+1)x)}{x}+b\frac{\tan x}{x}=(a+1)+b$. Continuity: $a+1+b=3$, so $a+b=2$.
Right limit: rationalise,
$\frac{\sqrt{ax+b^2x^2}-\sqrt{ax}}{b\sqrt a\,x\sqrt x}=\frac{b^2x^2}{b\sqrt a\,x\sqrt x\left(\sqrt{ax+b^2x^2}+\sqrt{ax}\right)}=\frac{b\sqrt x}{\sqrt a\cdot\sqrt x\left(\sqrt{a+b^2x}+\sqrt a\right)}\to\frac{b}{\sqrt a\cdot2\sqrt a}=\frac{b}{2a}.$
Continuity: $\frac{b}{2a}=3$, so $\frac{b}{a}=6$ (with $a+b=2$: $a=\frac27$, $b=\frac{12}{7}$, both positive).
Answer: (A). | ||||
| 10 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | Single correctEasy | |
A easy single correct question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $f'(x)=6x^2-18ax+12a^2=6(x-a)(x-2a)$.
With $a>0$, $f'$ changes from $+$ to $-$ at $x=a$ (local maximum) and from $-$ to $+$ at $x=2a$ (local minimum).
So $\alpha=a$ and $\alpha^2=2a$: $a^2=2a$, $a=2$. Then $\alpha=2$, $\alpha^2=4$.
The equation with roots 2 and 4: $x^2-6x+8=0$.
Answer: (C). | ||||
| 11 | Integral CalculusFundamental theorem of calculus | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on fundamental theorem of calculus, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Put $t=\sqrt{e^x-1}$: $e^x=1+t^2$, $dx=\frac{2t\,dt}{1+t^2}$, so $\int\frac{dx}{\sqrt{e^x-1}}=\int\frac{2\,dt}{1+t^2}=2\tan^{-1}t$.
At $x=\log_e4$: $t=\sqrt3$, $\tan^{-1}\sqrt3=\frac{\pi}{3}$.
$2\left(\frac{\pi}{3}-\tan^{-1}\sqrt{e^\alpha-1}\right)=\frac{\pi}{6}$, so $\tan^{-1}\sqrt{e^\alpha-1}=\frac{\pi}{4}$, $e^\alpha-1=1$, $e^\alpha=2$.
Roots $2$ and $\frac12$: sum $\frac52$, product 1, so $2x^2-5x+2=0$.
Answer: (C). | ||||
| 12 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Intersection: $x^2+2x-8=0$, $x=2$, $y=2$.
Outside the parabola means $y^2>2x$, i.e. the part to the left of the parabola. Integrate in $y$:
For $0\le y\le2$: $0\le x\le\frac{y^2}{2}$, area $\int_0^2\frac{y^2}{2}dy=\frac43$.
For $2\le y\le2\sqrt2$: $0\le x\le\sqrt{8-y^2}$, area $\int_2^{2\sqrt2}\sqrt{8-y^2}\,dy=\left[\frac{y}{2}\sqrt{8-y^2}+4\sin^{-1}\frac{y}{2\sqrt2}\right]_2^{2\sqrt2}=2\pi-(2+\pi)=\pi-2$.
Total $=\frac43+\pi-2=\pi-\frac23$.
Check: the quarter circle has area $2\pi$; the part inside the parabola is $\int_0^2\left(\sqrt{8-y^2}-\frac{y^2}{2}\right)dy=\pi+2-\frac43=\pi+\frac23$, and $2\pi-\left(\pi+\frac23\right)=\pi-\frac23$.
Answer: (A). | ||||
| 13 | Differential EquationsLinear differential equations | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations Divide by $\cos y$: $\sec^2y\frac{dy}{dx}+2x\tan y=x^3$.
Put $t=\tan y$: $\frac{dt}{dx}+2xt=x^3$, a linear equation with integrating factor $e^{x^2}$.
$te^{x^2}=\int x^3e^{x^2}dx=\frac12(x^2-1)e^{x^2}+C$ (substitute $u=x^2$).
$y(1)=0$ means $t=0$ at $x=1$: $0=\frac12(1-1)e+C$, so $C=0$.
$\tan y=\frac{x^2-1}{2}$. At $x=\sqrt3$: $\tan y=1$, $y=\frac{\pi}{4}$.
Answer: (C). | ||||
| 14 | Straight LinesSlope, parallel and perpendicular lines | Class 11: Straight Lines | Single correctMedium | |
A medium single correct question on slope, parallel and perpendicular lines, from Straight Lines. NCERT: Class 11: Straight Lines The lines from the origin have slopes $m_1=\frac25$ and $m_2=\frac a2$. The angle between them is $\frac{\pi}{4}$:
$\left|\frac{\frac a2-\frac25}{1+\frac a5}\right|=1.$
Case $+1$: $\frac a2-\frac25=1+\frac a5$, $\frac{3a}{10}=\frac75$, $a=\frac{14}{3}$.
Case $-1$: $\frac a2-\frac25=-1-\frac a5$, $\frac{7a}{10}=-\frac35$, $a=-\frac67$.
Both give an acute angle at the origin (the dot products $10+2a$ are positive), so the angle is exactly $\frac{\pi}{4}$ in each case.
Product $=\frac{14}{3}\times\left(-\frac67\right)=-4$; absolute value 4.
Answer: (D). | ||||
| 15 | Straight LinesDistance of a point from a line | Class 11: Straight Lines | Single correctEasy | |
A easy single correct question on distance of a point from a line, from Straight Lines. NCERT: Class 11: Straight Lines Image of $(x_1,y_1)$ in $x+2y-2=0$: $\frac{x-x_1}{1}=\frac{y-y_1}{2}=\frac{-2(x_1+2y_1-2)}{1^2+2^2}$.
Here $x_1+2y_1-2=-4+10-2=4$, so the ratio is $-\frac85$.
Image: $x=-4-\frac85=-\frac{28}{5}$, $y=5-\frac{16}{5}=\frac95$.
$r^2=\left(-\frac{28}{5}+4\right)^2+\left(\frac95-3\right)^2=\frac{64}{25}+\frac{36}{25}=4$, so $r=2$.
Answer: (B). | ||||
| 16 | Three Dimensional GeometrySkew lines and shortest distance | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on skew lines and shortest distance, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry The directions $(2,3,4)$ and $(4,6,8)$ are parallel, so the lines are parallel. Use $\vec d=(2,3,4)$, $|\vec d|=\sqrt{29}$.
Points: $P=(\lambda,4,3)$, $Q=(2,4,7)$; $\vec{PQ}=(2-\lambda,0,4)$.
Distance $=\frac{|\vec{PQ}\times\vec d|}{|\vec d|}$, with $\vec{PQ}\times\vec d=(0\cdot4-4\cdot3,\;4\cdot2-(2-\lambda)4,\;(2-\lambda)3-0)=(-12,\,4\lambda,\,6-3\lambda)$.
$|\vec{PQ}\times\vec d|^2=144+16\lambda^2+(6-3\lambda)^2=25\lambda^2-36\lambda+180$.
Set $\frac{25\lambda^2-36\lambda+180}{29}=\frac{169}{29}$: $25\lambda^2-36\lambda+11=0$, $(\lambda-1)(25\lambda-11)=0$.
$\lambda=1$ or $\frac{11}{25}$; of these only 1 is offered.
Answer: (C). | ||||
| 17 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Single correctMedium | |
A medium single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $\vec c\times(-2\vec a+3\vec b)=-(-2\vec a+3\vec b)\times\vec c$, so
$(\vec a+\vec b)\times\vec c+(-2\vec a+3\vec b)\times\vec c=\vec0$, i.e. $(-\vec a+4\vec b)\times\vec c=\vec0$.
Hence $\vec c=t(-\vec a+4\vec b)=t(40\hat i-3\hat j+3\hat k)$.
$2\vec a+3\vec b=41\hat i-5\hat j+5\hat k$, and $(2\vec a+3\vec b)\cdot\vec c=t(1640+15+15)=1670t=1670$, so $t=1$.
$|\vec c|^2=40^2+3^2+3^2=1618$.
Answer: (C). | ||||
| 18 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Single correctEasy | |
A easy single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $\vec b+\vec c=5\hat i+2\hat j+(\lambda-5)\hat k$, and $\vec r=\frac{\vec b+\vec c}{|\vec b+\vec c|}$.
$\vec r\cdot\vec a=\frac{5+4+3(\lambda-5)}{\sqrt{25+4+(\lambda-5)^2}}=\frac{3\lambda-6}{\sqrt{29+(\lambda-5)^2}}=3.$
So $(\lambda-2)^2=29+(\lambda-5)^2$: $-4\lambda+4=29-10\lambda+25$, $6\lambda=50$, $\lambda=\frac{25}{3}$ (and $3\lambda-6=19>0$, as needed).
$3\lambda=25$.
Answer: (A). | ||||
| 19 | ProbabilityBayes' theorem | Class 11: Probability; Class 12: Probability | Single correctEasy | |
A easy single correct question on bayes' theorem, from Probability. NCERT: Class 11: Probability; Class 12: Probability Each bag is chosen with probability $\frac13$. Probability of a one-rupee coin: from X $\frac59$, from Y $\frac49$, from Z $\frac39$.
By Bayes' theorem:
$P(Y\mid\text{one-rupee})=\frac{\frac13\cdot\frac49}{\frac13\left(\frac59+\frac49+\frac39\right)}=\frac{4}{12}=\frac13.$
Answer: (B). | ||||
| 20 | TrigonometryTrigonometric functions | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on trigonometric functions, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions $\cos36^\circ=\frac{\sqrt5+1}{4}$, $\sin18^\circ=\frac{\sqrt5-1}{4}$.
Numerator $\times4$: $3(\sqrt5+1)+5(\sqrt5-1)=8\sqrt5-2$.
Denominator $\times4$: $5(\sqrt5+1)-3(\sqrt5-1)=2\sqrt5+8$.
Value $=\frac{8\sqrt5-2}{2\sqrt5+8}=\frac{4\sqrt5-1}{\sqrt5+4}=\frac{(4\sqrt5-1)(4-\sqrt5)}{16-5}=\frac{16\sqrt5-20-4+\sqrt5}{11}=\frac{17\sqrt5-24}{11}.$
So $a=17$, $b=24$, $c=11$ ($\gcd(17,11)=1$), and $a+b+c=52$.
Answer: (C). | ||||
| 21 | Quadratic EquationsRoots in real and complex systems | Class 11: Complex Numbers and Quadratic Equations | NumericalMedium | |
A medium numerical question on roots in real and complex systems, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations Put $t=x+2$. Then $|x+1||x+3|=|t-1||t+1|=|t^2-1|$, and the equation is $|t^2-1|-4|t|+5=0$. Let $u=|t|\ge0$.
If $u\ge1$: $u^2-1-4u+5=(u-2)^2=0$, so $u=2$.
If $0\le u<1$: $1-u^2-4u+5=0$, $u^2+4u-6=0$, $u=-2\pm\sqrt{10}$; neither lies in $[0,1)$ ($\sqrt{10}-2\approx1.16$).
So $|t|=2$, $t=\pm2$, $x=0$ or $x=-4$.
Check $x=0$: $1\cdot3-8+5=0$. $x=-4$: $3\cdot1-8+5=0$.
Two distinct real roots.
Answer: 2. | ||||
| 22 | Straight LinesVarious forms of the equation of a line | Class 11: Straight Lines | NumericalMedium | |
A medium numerical question on various forms of the equation of a line, from Straight Lines. NCERT: Class 11: Straight Lines The incident ray, produced beyond the mirror, passes through the image of $(7,2)$ in the line $2x+y-6=0$.
$2(7)+2-6=10$, so the image is $(7,2)-\frac{2\cdot10}{2^2+1^2}(2,1)=(7-8,\,2-4)=(-1,-2)$.
(Both $(3,10)$ and $(7,2)$ give $2x+y-6=10>0$, so they are on the same side of the mirror, as a reflection needs.)
Incident ray through $(3,10)$ and $(-1,-2)$: slope $\frac{12}{4}=3$, $y-10=3(x-3)$, i.e. $3x-y+1=0$.
So $a=3$, $b=-1$: $a^2+b^2+3ab=9+1-9=1$.
Answer: 1. | ||||
| 23 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | NumericalEasy | |
A easy numerical question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series The terms are $2,5,8,\dots$ with $a_n=2+3(n-1)$, and row $k$ holds $k$ terms.
Rows 1 to 9 hold $1+2+\dots+9=45$ terms, so row 10 holds $a_{46}$ to $a_{55}$.
$a_{46}=2+3\cdot45=137$, $a_{55}=2+3\cdot54=164$.
Sum $=\frac{10}{2}(137+164)=5\times301=1505$.
Answer: 1505. | ||||
| 24 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | NumericalMedium | |
A medium numerical question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives The region is symmetric about the $x$-axis, so take the rectangle with one side on $x=24$ and the opposite corners $(\frac{y^2}{2},\pm y)$ on the parabola, $0<y<\sqrt{48}$.
Area $S(y)=2y\left(24-\frac{y^2}{2}\right)=48y-y^3$.
$S'(y)=48-3y^2=0$ gives $y=4$, and $S''(4)=-24<0$, a maximum.
$S(4)=192-64=128$.
Answer: 128. | ||||
| 25 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalMedium | |
A medium numerical question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability $\alpha$: write $h=\sqrt{\tan x}-\sqrt x\to0$. Then $\frac{e^{\sqrt{\tan x}}-e^{\sqrt x}}{h}=e^{\sqrt x}\cdot\frac{e^{h}-1}{h}\to e^0\cdot1=1$. (For small $x>0$, $\tan x>x$, so $h>0$ and the expression is defined.) $\alpha=1$.
$\beta$: a $1^\infty$ form, $\beta=e^{\lim\frac12\cot x\cdot\sin x}=e^{\lim\frac12\cos x}=e^{1/2}=\sqrt e$.
Roots $1$ and $\sqrt e$ of $ax^2+bx-\sqrt e=0$:
Product $=-\frac{\sqrt e}{a}=\sqrt e$, so $a=-1$.
Sum $=-\frac ba=b=1+\sqrt e$.
$a+b=\sqrt e$, and $12\log_e\sqrt e=12\cdot\frac12=6$.
Answer: 6. | ||||
| 26 | Integral CalculusIntegration by substitution | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on integration by substitution, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Write the integrand as $\frac{1}{(x-1)^{4/5}(x+3)^{6/5}}=\left(\frac{x-1}{x+3}\right)^{-4/5}\frac{1}{(x+3)^2}$.
Put $t=\frac{x-1}{x+3}$: $dt=\frac{(x+3)-(x-1)}{(x+3)^2}dx=\frac{4}{(x+3)^2}dx$.
Integral $=\frac14\int t^{-4/5}dt=\frac14\cdot5t^{1/5}+C=\frac54\left(\frac{x-1}{x+3}\right)^{1/5}+C$.
Matching: $\alpha=1$, $\beta=1$, $A=\frac54$, $B=\frac15$.
$\alpha+\beta+20AB=2+20\cdot\frac54\cdot\frac15=2+5=7$.
Answer: 7. | ||||
| 27 | Differential EquationsSeparation of variables | Class 12: Differential Equations | NumericalMedium | |
A medium numerical question on separation of variables, from Differential Equations. NCERT: Class 12: Differential Equations Divide by $xy$: $\frac{dy}{y}-\frac{dx}{x}+(x\,dy+y\,dx)=0$, i.e. $d(\ln|y|)-d(\ln|x|)+d(xy)=0$.
Integrate: $\ln|y|-\ln|x|+xy=K$.
$y(1)=2$: $\ln2+2=K$.
So $\ln\frac{|y|}{|x|}=\ln2+2-xy$, i.e. $|y|e^{xy}=2e^2|x|$, or $|y|e^{xy-2}=2|x|$.
Compare with $\alpha|x|=|y|e^{xy-\beta}$: $\alpha=2e^{2-\beta}$. For natural numbers this holds only with $\beta=2$, $\alpha=2$.
$\alpha+\beta=4$.
Answer: 4. | ||||
| 28 | Conic SectionsHyperbola | Class 11: Conic Sections | NumericalMedium | |
A medium numerical question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections $a^2=3$, $b^2=5$: $e^2=1+\frac53=\frac83$, $ae=\sqrt{3\cdot\frac83}=2\sqrt2$. So $S=(2\sqrt2,0)$.
SAB is a diameter, so A is the midpoint of SB: $B=2A-S=(2\sqrt6-2\sqrt2,\;2\sqrt5)$.
Triangle OSB has base $OS=2\sqrt2$ on the $x$-axis and height $2\sqrt5$ (the $y$-coordinate of B).
Area $=\frac12\cdot2\sqrt2\cdot2\sqrt5=2\sqrt{10}$, and its square is 40.
Answer: 40. | ||||
| 29 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | NumericalMedium | |
A medium numerical question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry A point of the line: $F=(t,\,1+2t,\,2+3t)$. $\vec{FQ}=(1-t,\,5-2t,\,2-3t)$ must be perpendicular to $(1,2,3)$:
$(1-t)+2(5-2t)+3(2-3t)=17-14t=0$, $t=\frac{17}{14}$.
Foot $F=\left(\frac{17}{14},\frac{48}{14},\frac{79}{14}\right)$.
Image $P=2F-Q=\left(\frac{34}{14}-1,\frac{96}{14}-6,\frac{158}{14}-4\right)=\left(\frac{10}{7},\frac{6}{7},\frac{51}{7}\right)$.
$2\alpha+\beta+\gamma=\frac{20+6+51}{7}=\frac{77}{7}=11$.
Answer: 11. | ||||
| 30 | StatisticsStandard deviation and variance | Class 11: Statistics | NumericalMedium | |
A medium numerical question on standard deviation and variance, from Statistics. NCERT: Class 11: Statistics Mean 18: $9+25+a+b+c=90$, so $a+b+c=56$.
Let $p=a-18$, $q=b-18$, $r=c-18$. Then $p+q+r=56-54=2$.
Mean deviation 4: $|9-18|+|25-18|+|p|+|q|+|r|=20$, so $|p|+|q|+|r|=20-16=4$.
Variance $\frac{136}{5}$: $\sum(x_i-18)^2=136$, so $81+49+p^2+q^2+r^2=136$, $p^2+q^2+r^2=6$.
Integers with squares summing to 6: $\{|p|,|q|,|r|\}=\{1,1,2\}$ (absolute sum 4, as required). With sum 2 and $p<q<r$: $p=-1$, $q=1$, $r=2$.
So $a=17$, $b=19$, $c=20$, and $2a+b-c=34+19-20=33$.
Answer: 33. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 31 | Physical World and MeasurementUnits and dimensions | Class 11: Units and Measurement | Single correctEasy | |
A easy single correct question on units and dimensions, from Physical World and Measurement. NCERT: Class 11: Units and Measurement The energy density of an electric field is $u=\frac12\epsilon_0E^2$, so $\epsilon_0E^2$ has the dimensions of energy per unit volume:
$\frac{[\mathrm{ML^2T^{-2}}]}{[\mathrm{L^3}]}=[\mathrm{ML^{-1}T^{-2}}]$.
(The same as pressure.)
Answer: (D). | ||||
| 32 | KinematicsProjectile motion | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on projectile motion, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane $R=\frac{u^2\sin2\theta}{g}=\frac{2u^2\sin\theta\cos\theta}{g}$ and $H=\frac{u^2\sin^2\theta}{2g}$.
$R=H$: $2\sin\theta\cos\theta=\frac{\sin^2\theta}{2}$, so $\tan\theta=4$, $\theta=\tan^{-1}(4)$.
Answer: (B). | ||||
| 33 | Laws of MotionFriction | Class 11: Laws of Motion | Single correctMedium | |
A medium single correct question on friction, from Laws of Motion. NCERT: Class 11: Laws of Motion Same length $s$ from rest: $s=\frac12at^2$, so $t\propto\frac{1}{\sqrt a}$.
Smooth: $a_1=g\sin45^\circ$. Rough: $a_2=g(\sin45^\circ-\mu\cos45^\circ)=g\sin45^\circ(1-\mu)$.
$\frac{t_2}{t_1}=n=\sqrt{\frac{a_1}{a_2}}=\frac{1}{\sqrt{1-\mu}}$, so $1-\mu=\frac{1}{n^2}$ and $\mu=1-\frac{1}{n^2}$.
Answer: (B). | ||||
| 34 | Work, Energy and PowerWork-energy theorem | Class 11: Work, Energy and Power | Single correctMedium | |
A medium single correct question on work-energy theorem, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power From the figure: mass 5 kg, smooth incline of length 10 m at $30^\circ$, then a rough horizontal strip of 2 m ($\mu=0.5$) before the spring ($k=100$ N/m). Take $g=10$ m/s$^2$.
Height of the top: $h=10\sin30^\circ=5$ m, so the energy released is $mgh=5\times10\times5=250$ J.
Work against friction over the 2 m strip: $\mu mg\,d=0.5\times5\times10\times2=50$ J.
Kinetic energy on reaching the spring: $250-50=200$ J.
The rest of the floor is smooth, so all of it is stored in the spring: $\frac12kx^2=200$, $50x^2=200$, $x=2$ m.
Answer: (D). | ||||
| 35 | Motion of System of Particles and Rigid BodyAngular momentum | Class 11: Systems of Particles and Rotational Motion | Single correctEasy | |
A easy single correct question on angular momentum, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion $I_1=\frac12MR^2$, $I_2=\frac12\cdot\frac M2R^2=\frac14MR^2$.
No external torque about the axis, so angular momentum is conserved: $I_1\omega=(I_1+I_2)\omega'$.
$\omega'=\frac{\frac12}{\frac12+\frac14}\omega=\frac23\omega$.
Answer: (B). | ||||
| 36 | GravitationOrbital motion | Class 11: Gravitation | Single correctEasy | |
A easy single correct question on orbital motion, from Gravitation. NCERT: Class 11: Gravitation Orbital speed $v_o=\sqrt{\frac{GM}{r}}$, so $v_o\propto\frac{1}{\sqrt r}$.
$\frac{v_B}{v_A}=\sqrt{\frac{4R}{R}}=2$, so $v_B=2\times3v=6v$.
Answer: (C). | ||||
| 37 | Properties of Bulk MatterFluid mechanics | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctEasy | |
A easy single correct question on fluid mechanics, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter As in the figure, the cube is wholly submerged, part in the water below and part in the kerosene above. Let $V_w$ be the volume in water and $V_k$ in kerosene, $V_w+V_k=V$.
Floating: weight = total buoyancy, $0.9V\rho g=1\cdot V_w\rho g+0.8V_k\rho g$ ($\rho$ = density of water).
$0.9(V_w+V_k)=V_w+0.8V_k$, so $0.1V_k=0.1V_w$ and $V_w:V_k=1:1$.
Answer: (D). | ||||
| 38 | ThermodynamicsFirst law | Class 11: Thermodynamics | Single correctEasy | |
A easy single correct question on first law, from Thermodynamics. NCERT: Class 11: Thermodynamics Isobaric: $W=nR\Delta T$ and $Q=nC_p\Delta T$.
$\frac QW=\frac{C_p}{R}=\frac{\gamma}{\gamma-1}=\frac{1.4}{0.4}=3.5$.
$Q=3.5\times100=350$ J.
Answer: (B). | ||||
| 39 | Behaviour of Perfect Gas and Kinetic TheoryMean free path | Class 11: Kinetic Theory | StatementsEasy | |
A easy statements question on mean free path, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory Statement I: $\lambda=\frac{1}{\sqrt2\pi d^2n}$, so the mean free path varies as $\frac{1}{d^2}$. True.
Statement II: the average kinetic energy of a molecule is $\frac32k_BT$, proportional to the absolute temperature. True.
Answer: (A). | ||||
| 40 | ElectrostaticsConductors and dielectrics | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | Single correctMedium | |
A medium single correct question on conductors and dielectrics, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance Before: $C=\frac{\epsilon_0A}{d}$ with $d=0.6$ cm.
With the slab (thickness $t=0.6$ cm) the plate is moved out to a separation $d'=0.8$ cm: $C'=\frac{\epsilon_0A}{d'-t+\frac tK}$.
Equal capacitances: $0.8-0.6+\frac{0.6}{K}=0.6$, so $\frac{0.6}{K}=0.4$ and $K=1.5$.
(The value of $\epsilon_0$ is not needed.)
Answer: (A). | ||||
| 41 | Current ElectricityResistivity | Class 12: Current Electricity | Single correctEasy | |
A easy single correct question on resistivity, from Current Electricity. NCERT: Class 12: Current Electricity The heat needed is fixed: $H=\frac{V^2}{R}t$ with the same $V$, so $t\propto R\propto L$ (same wire material and cross-section).
$\frac{L'}{L}=\frac{15}{20}=\frac34$: the length must be decreased to $\frac34$ of its initial length.
Answer: (B). | ||||
| 42 | Magnetic Effects of Current and MagnetismAmpere's law | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | Single correctEasy | |
A easy single correct question on ampere's law, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter By Ampere's law:
Inside ($r<a$): $B=\frac{\mu_0Ir}{2\pi a^2}$; at $r=\frac a2$: $B_1=\frac{\mu_0I}{4\pi a}$.
Outside ($r>a$): $B=\frac{\mu_0I}{2\pi r}$; at $r=2a$: $B_2=\frac{\mu_0I}{4\pi a}$.
$B_1:B_2=1:1$.
Answer: (C). | ||||
| 43 | Electromagnetic Induction and Alternating CurrentsAC circuits | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctMedium | |
A medium single correct question on ac circuits, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current Current: $I=\frac{36}{90}=0.4$ A.
The voltages across R and L are in quadrature: $V_L=\sqrt{120^2-36^2}=\sqrt{13104}\approx114.5$ V.
$X_L=\frac{V_L}{I}=\frac{114.5}{0.4}\approx286$ $\Omega$.
$L=\frac{X_L}{2\pi f}=\frac{286}{2\pi\times60}=\frac{286}{377}\approx0.76$ H.
Answer: (D). | ||||
| 44 | Oscillations and WavesWave motion | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on wave motion, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves Compare with $y=A\cos(\omega t-kx)$: $\omega=2\pi\times330$ rad/s.
$f=\frac{\omega}{2\pi}=330$ Hz.
Answer: (C). | ||||
| 45 | OpticsLenses and mirrors | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | Single correctMedium | |
A medium single correct question on lenses and mirrors, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics From the figure: object 30 cm to the left of lens 1 ($f_1=10$ cm); lens 2 ($f_2=-10$ cm) 5 cm after lens 1; lens 3 ($f_3=30$ cm) 10 cm after lens 2. Use $\frac1v-\frac1u=\frac1f$.
Lens 1: $u=-30$, $\frac1v=\frac1{10}-\frac1{30}=\frac1{15}$, $v=+15$ cm, i.e. 10 cm beyond lens 2.
Lens 2: the image of lens 1 is a virtual object at $u=+10$ cm. $\frac1v=-\frac1{10}+\frac1{10}=0$, so the rays leave lens 2 parallel to the axis.
Lens 3: parallel rays focus at its focal point, 30 cm to the right of the third lens.
Answer: (D). | ||||
| 46 | Dual Nature of Matter and Radiationde Broglie wavelength | Class 12: Dual Nature of Radiation and Matter | Single correctEasy | |
A easy single correct question on de broglie wavelength, from Dual Nature of Matter and Radiation. NCERT: Class 12: Dual Nature of Radiation and Matter $\lambda=\frac hp$: equal wavelengths mean equal momenta.
$K=\frac{p^2}{2m}$, so with the same $p$ the heavier particle has less kinetic energy. $m_p>m_e$, hence $K_p<K_e$.
Answer: (B). | ||||
| 47 | Atoms and NucleiNuclear binding energy | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on nuclear binding energy, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei ${}_5^{12}\mathrm B$ has 5 protons and $12-5=7$ neutrons.
Mass defect $\Delta m=5M_p+7M_n-M_0$, and the binding energy is $\Delta m\,c^2=(5M_p+7M_n-M_0)c^2$.
Answer: (C). | ||||
| 48 | Atoms and NucleiNuclear fission and fusion | Class 12: Atoms; Class 12: Nuclei | Single correctEasy | |
A easy single correct question on nuclear fission and fusion, from Atoms and Nuclei. NCERT: Class 12: Atoms; Class 12: Nuclei Mass number: $236=141+92+3A$, so $A=1$.
Charge number: $92=56+36+3Z$, so $Z=0$.
A particle with $A=1$, $Z=0$ is a neutron.
Answer: (C). | ||||
| 49 | Experimental SkillsVernier calipers and screw gauge | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on vernier calipers and screw gauge, from Experimental Skills. NCERT: No matching NCERT chapter Least count $=\frac{1\text{ mm}}{100}=0.01$ mm.
Zero of the circular scale 5 divisions below the reference line: positive zero error $=+5\times0.01=+0.05$ mm.
Observed reading $=4\times1+60\times0.01=4.60$ mm.
Corrected diameter $=4.60-0.05=4.55$ mm.
Answer: (D). | ||||
| 50 | Experimental SkillsVernier calipers and screw gauge | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on vernier calipers and screw gauge, from Experimental Skills. NCERT: No matching NCERT chapter Least count $=1$ MSD $-$ 1 VSD $=\frac{1}{20N}$ cm $=\frac{1}{2N}$ mm.
With 1 MSD = 1 mm: 1 VSD $=1-\frac{1}{2N}=\frac{2N-1}{2N}$ mm.
N VSD $=N\cdot\frac{2N-1}{2N}=\frac{2N-1}{2}$ mm, which is $\frac{2N-1}{2}$ main-scale divisions.
Answer: (B). | ||||
| 51 | KinematicsProjectile motion | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | NumericalEasy | |
A easy numerical question on projectile motion, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane A body thrown horizontally from height $h$ falls for $t=\sqrt{\frac{2h}{g}}$ and lands at $x=ut$, independent of its mass.
First: $v\sqrt{\frac{2H}{g}}=100$ m.
Second: $\frac v2\sqrt{\frac{2(4H)}{g}}=\frac v2\cdot2\sqrt{\frac{2H}{g}}=v\sqrt{\frac{2H}{g}}=100$ m.
Answer: 100. | ||||
| 52 | Laws of MotionCircular motion dynamics | Class 11: Laws of Motion | NumericalHard | |
A hard numerical question on circular motion dynamics, from Laws of Motion. NCERT: Class 11: Laws of Motion In the frame of the table the only radial force on the ball is the centrifugal force $m\omega^2r$ (the groove is smooth; its normal force is sideways).
$v\frac{dv}{dr}=\omega^2r$, so $\int_0^v v\,dv=\omega^2\int_1^3r\,dr$: $\frac{v^2}{2}=\frac{\omega^2}{2}(9-1)$.
$v=\omega\sqrt8=2\sqrt2\,\omega$ m/s, so $x=2$.
Answer: 2. | ||||
| 53 | Properties of Bulk MatterViscosity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | NumericalEasy | |
A easy numerical question on viscosity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Volume is conserved: $\frac43\pi R^3=8\cdot\frac43\pi r^3$, so $R=2r$.
Terminal velocity (Stokes' law) $v_t=\frac{2r^2(\rho-\sigma)g}{9\eta}\propto r^2$.
$v'=10\times2^2=40$ cm/s.
Answer: 40. | ||||
| 54 | Oscillations and WavesSHM | Class 11: Oscillations; Class 11: Waves | NumericalEasy | |
A easy numerical question on shm, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves $\omega=2\pi f=2\pi\cdot\frac{25}{\pi}=50$ rad/s.
Total energy $E=0.5+0.4=0.9$ J $=\frac12m\omega^2A^2=\frac12(0.2)(2500)A^2=250A^2$.
$A^2=0.0036$, $A=0.06$ m $=6$ cm.
Check: at $x=0.04$ m, $U=\frac12(0.2)(2500)(0.0016)=0.4$ J, as given.
Answer: 6. | ||||
| 55 | ElectrostaticsCoulomb's law | Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance | NumericalMedium | |
A medium numerical question on coulomb's law, from Electrostatics. NCERT: Class 12: Electric Charges and Fields; Class 12: Electrostatic Potential and Capacitance From the figure, put the foot of the 4 cm perpendicular at the origin: $P=(0,4)$, $+q_2$ at $(-2,0)$, $-q_3$ at $(3,0)$ (cm).
$q_2P=\sqrt{4+16}=\sqrt{20}$, $q_3P=\sqrt{9+16}=5$.
Field of $+q_2$ at P points away from $q_2$; its Y component is $\frac{kq_2}{20}\cdot\frac{4}{\sqrt{20}}$ (upward).
Field of $-q_3$ at P points toward $q_3$; its Y component is $\frac{k|q_3|}{25}\cdot\frac45$ (downward).
Net Y component zero: $\frac{4q_2}{20\sqrt{20}}=\frac{4|q_3|}{125}$, so $\left|\frac{q_2}{q_3}\right|=\frac{20\sqrt{20}}{125}=\frac{40\sqrt5}{125}=\frac{8\sqrt5}{25}=\frac{8}{5\sqrt5}$.
So $x=5$.
Answer: 5. | ||||
| 56 | Current ElectricityOhm's law | Class 12: Current Electricity | NumericalMedium | |
A medium numerical question on ohm's law, from Current Electricity. NCERT: Class 12: Current Electricity Heater resistance: $R_h=\frac{V^2}{P}=\frac{100^2}{1000}=10$ $\Omega$.
At 62.5 W: $\frac{V_h^2}{10}=62.5$, so $V_h=25$ V across BC.
The 10 $\Omega$ resistor then carries the rest: $100-25=75$ V, current $7.5$ A.
The heater and R in parallel take 7.5 A at 25 V: $R_{BC}=\frac{25}{7.5}=\frac{10}{3}$ $\Omega$.
$\frac1R+\frac1{10}=\frac3{10}$, so $R=5$ $\Omega$.
Answer: 5. | ||||
| 57 | Magnetic Effects of Current and MagnetismMagnetic properties of materials | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalEasy | |
A easy numerical question on magnetic properties of materials, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter The magnet is demagnetised when the solenoid's field intensity equals the coercivity: $H=nI$.
$n=\frac{150}{0.30}=500$ turns/m.
$I=\frac Hn=\frac{5\times10^3}{500}=10$ A.
Answer: 10. | ||||
| 58 | Electromagnetic Induction and Alternating CurrentsAC circuits | Class 12: Electromagnetic Induction; Class 12: Alternating Current | NumericalEasy | |
A easy numerical question on ac circuits, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current $E_{rms}=\frac{110\sqrt2}{\sqrt2}=110$ V, $\omega=100$ rad/s.
$X_C=\frac{1}{\omega C}=\frac{1}{100\times2\times10^{-6}}=5000$ $\Omega$.
$I_{rms}=\frac{110}{5000}=0.022$ A $=22$ mA.
Answer: 22. | ||||
| 59 | OpticsInterference and diffraction | Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics | NumericalMedium | |
A medium numerical question on interference and diffraction, from Optics. NCERT: Class 12: Ray Optics and Optical Instruments; Class 12: Wave Optics Fringe width of the double-slit pattern: $\beta=\frac{\lambda D}{d}$.
Width of the single-slit central maximum: $\frac{2\lambda D}{a}$.
Ten fringes within it: $\frac{2\lambda D}{a}=10\frac{\lambda D}{d}$, so $a=\frac{d}{5}=\frac{1\text{ mm}}{5}=0.2$ mm $=2\times10^{-4}$ m.
Answer: 2. | ||||
| 60 | Electronic DevicesSpecial-purpose diodes | Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits | NumericalMedium | |
A medium numerical question on special-purpose diodes, from Electronic Devices. NCERT: Class 12: Semiconductor Electronics: Materials, Devices and Simple Circuits In the figure the LED (forward biased) and the zener (reverse biased) are in series across the part PQ of the wire.
To just glow the LED, the voltage across PQ must be $1.8+3.2=5$ V.
Potential gradient along the wire: $\frac{20\text{ V}}{20\text{ cm}}=1$ V/cm.
$PQ=\frac{5}{1}=5$ cm.
Answer: 5. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 61 | Chemical Bonding and Molecular StructureMolecular orbital theory | Class 11: Chemical Bonding and Molecular Structure | Single correctEasy | |
A easy single correct question on molecular orbital theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure By the LCAO method two atomic orbitals combine in two ways.
Additive overlap, $\psi_A+\psi_B$, gives the bonding orbital $\sigma$.
Subtractive overlap, $\psi_A-\psi_B$, gives the antibonding orbital $\sigma^*$, with a node between the nuclei.
(Equal coefficients for identical atoms; $\psi_A\pm2\psi_B$ are not the standard combinations.)
Answer: (B). | ||||
| 62 | EquilibriumpH and buffers | Class 11: Equilibrium | StatementsEasy | |
A easy statements question on ph and buffers, from Equilibrium. NCERT: Class 11: Equilibrium Statement I: a buffer is a weak acid with its salt (or a weak base with its salt) in suitable proportions; the acid or base must be weak and the amounts are not arbitrary (buffer action is best when the ratio is near 1). 'Any particular quantities' of 'an acid or a base' is wrong. False.
Statement II: the pH of blood (about 7.4) is held mainly by the $\mathrm{H_2CO_3/HCO_3^-}$ buffer. True.
Answer: (D). | ||||
| 63 | ElectrochemistryNernst equation | Class 12: Electrochemistry | Single correctMedium | |
A medium single correct question on nernst equation, from Electrochemistry. NCERT: Class 12: Electrochemistry Cell reaction: $2\mathrm{Tl}+\mathrm{Cu^{2+}}\to2\mathrm{Tl^+}+\mathrm{Cu}$ ($n=2$).
Nernst equation: $E=E^\circ-\frac{0.059}{2}\log\frac{[\mathrm{Tl^+}]^2}{[\mathrm{Cu^{2+}}]}$.
E rises when the quotient falls: by raising $[\mathrm{Cu^{2+}}]$ or lowering $[\mathrm{Tl^+}]$.
Raising $[\mathrm{Tl^+}]$ lowers E; changing both together does not fix the direction (the $\mathrm{Tl^+}$ term is squared). Of the options only increasing $[\mathrm{Cu^{2+}}]$ surely increases the emf.
Answer: (B). | ||||
| 64 | ElectrochemistryElectrochemical cells | Class 12: Electrochemistry | Single correctEasy | |
A easy single correct question on electrochemical cells, from Electrochemistry. NCERT: Class 12: Electrochemistry Anode (oxidation): $\frac12\mathrm{H_2}\to\mathrm{H^+}+e^-$, a hydrogen electrode $\mathrm{Pt|H_2}$ dipping in an acid.
Cathode (reduction): $\mathrm{AgCl}+e^-\to\mathrm{Ag}+\mathrm{Cl^-}$, a silver–silver chloride electrode.
Both electrodes share one electrolyte that holds the $\mathrm{H^+}$ and $\mathrm{Cl^-}$ produced, HCl solution: $\mathrm{Pt|H_2(g)|HCl(soln.)|AgCl(s)|Ag}$.
With $\mathrm{AgNO_3}$ the cathode reaction would be $\mathrm{Ag^+}+e^-\to\mathrm{Ag}$, and with KCl the anode would not be in an acid solution producing $\mathrm{H^+}$ and $\mathrm{Cl^-}$ as written.
Answer: (C). | ||||
| 65 | Chemical KineticsRate laws | Class 12: Chemical Kinetics | Single correctMedium | |
A medium single correct question on rate laws, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics Net rate of formation of B: $\frac{d[\mathrm B]}{dt}=K_1[\mathrm A]-K_2[\mathrm B]$.
Setting it to zero (steady state): $K_1[\mathrm A]=K_2[\mathrm B]$, so $[\mathrm B]=\frac{K_1}{K_2}[\mathrm A]$.
Answer: (D). | ||||
| 66 | Classification of Elements and PeriodicityPeriodic trends | Class 11: Classification of Elements and Periodicity in Properties | Single correctEasy | |
A easy single correct question on periodic trends, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties (A) True: non-metals attract electrons more strongly; they are more electronegative than metals.
(B) False: non-metals have higher ionisation enthalpies than metals.
(C) True: a highly reactive metal and a highly reactive non-metal (e.g. Na and Cl) form ionic compounds.
(D) False: non-metal oxides are generally acidic (or neutral).
(E) False: metal oxides are generally basic (some amphoteric).
Correct: (A) and (C) only.
Answer: (A). | ||||
| 67 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter (A) Correct: $\mathrm{N_2}$ is very unreactive at room temperature because of its strong triple bond.
(B) Correct: $-3$, $+3$ and $+5$ are the common oxidation states.
(C) Correct: small size and high electronegativity let nitrogen form $p\pi$–$p\pi$ multiple bonds.
(D) Incorrect: because of the inert pair effect the stability of $+5$ decreases (and of $+3$ increases) down the group; $\mathrm{Bi(V)}$ is a strong oxidant.
(E) Incorrect: nitrogen has no d orbitals; its maximum covalency is 4.
Incorrect statements: (D) and (E) only.
Answer: (C). | ||||
| 68 | d- and f-Block ElementsPotassium dichromate and permanganate | Class 12: The d- and f-Block Elements | Single correctEasy | |
A easy single correct question on potassium dichromate and permanganate, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements The interconversion is $\mathrm{Cr_2O_7^{2-}}+2\mathrm{OH^-}\rightleftharpoons2\mathrm{CrO_4^{2-}}+\mathrm{H_2O}$ (equivalently $2\mathrm{CrO_4^{2-}}+2\mathrm{H^+}\rightleftharpoons\mathrm{Cr_2O_7^{2-}}+\mathrm{H_2O}$).
Adding base removes $\mathrm{H^+}$ / supplies $\mathrm{OH^-}$, so the equilibrium moves to chromate (right); acid moves it to dichromate.
Answer: (B). | ||||
| 69 | d- and f-Block ElementsPotassium dichromate and permanganate | Class 12: The d- and f-Block Elements | StatementsEasy | |
A easy statements question on potassium dichromate and permanganate, from d- and f-Block Elements. NCERT: Class 12: The d- and f-Block Elements Statement I: $2\mathrm{MnO_2}+4\mathrm{KOH}+\mathrm{O_2}\to2\mathrm{K_2MnO_4}+2\mathrm{H_2O}$; potassium manganate is dark green. True.
Statement II: in the manufacture of $\mathrm{KMnO_4}$, manganate is oxidised electrolytically in alkaline solution: $\mathrm{MnO_4^{2-}}\to\mathrm{MnO_4^-}+e^-$. True.
Answer: (A). | ||||
| 70 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | Matching listMedium | |
A medium matching list question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Spin-only moment $\mu=\sqrt{n(n+2)}$ B.M. for $n$ unpaired electrons.
(A) $\mathrm{Cr^{3+}}$, $3d^3$: $n=3$, $\mu=\sqrt{15}=3.87$ → (II).
(B) $[\mathrm{NiCl_4}]^{2-}$: $\mathrm{Ni^{2+}}$, $3d^8$, weak-field tetrahedral, $n=2$, $\mu=\sqrt8=2.83$ → (IV).
(C) $[\mathrm{CoF_6}]^{3-}$: $\mathrm{Co^{3+}}$, $3d^6$, weak-field $\mathrm{F^-}$, high spin, $n=4$, $\mu=\sqrt{24}=4.90$ → (I).
(D) $[\mathrm{Ni(CN)_4}]^{2-}$: $\mathrm{Ni^{2+}}$, $3d^8$, strong-field square planar ($dsp^2$), $n=0$ → (III).
A→II, B→IV, C→I, D→III.
Answer: (B). | ||||
| 71 | Organic Chemistry: Some Basic Principles and TechniquesQualitative and quantitative analysis | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on qualitative and quantitative analysis, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques The oxidising agent (e.g. $\mathrm{Na_2O_2}$) converts phosphorus to phosphate. Boiled with nitric acid and treated with ammonium molybdate, phosphate gives a yellow precipitate of ammonium phosphomolybdate:
$\mathrm{H_3PO_4}+12(\mathrm{NH_4})_2\mathrm{MoO_4}+21\mathrm{HNO_3}\to(\mathrm{NH_4})_3\mathrm{PO_4\cdot12MoO_3}+21\mathrm{NH_4NO_3}+12\mathrm{H_2O}$.
Answer: (B). | ||||
| 72 | Organic Chemistry: Some Basic Principles and TechniquesQualitative and quantitative analysis | Class 11: Organic Chemistry – Some Basic Principles and Techniques | StatementsEasy | |
A easy statements question on qualitative and quantitative analysis, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Kjeldahl's method works by converting the nitrogen of the compound into ammonium sulphate on digestion with conc. $\mathrm{H_2SO_4}$.
Nitrogen in a ring (pyridine), in nitro and azo groups is not converted into ammonium sulphate under these conditions, so the method does not apply to pyridine.
Statement I: false. Statement II: false.
Answer: (B). | ||||
| 73 | Organic Chemistry: Some Basic Principles and TechniquesReaction intermediates | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on reaction intermediates, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques The positively charged carbon of a carbocation has three bonds and no lone pair; it is $sp^2$ hybridised, with the empty p orbital perpendicular to the plane of the three bonds. Its shape is trigonal planar.
Answer: (B). | ||||
| 74 | Organic Chemistry: Some Basic Principles and TechniquesIUPAC nomenclature | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on iupac nomenclature, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques The structure is $\mathrm{CH_3-CH(CH_3)-CH_2-CH_2-CH(CH_3)-CH(CH_3)-CH_2-CH_3}$.
The longest chain has 8 carbons (octane), carrying three methyl groups.
Numbered from the left the methyls are at 2, 5, 6; from the right at 3, 4, 7. The first point of difference favours the left: 2, 5, 6.
Name: 2,5,6-trimethyloctane.
(The two 'ethyl…heptane' names use a shorter chain and are wrong.)
Answer: (C). | ||||
| 75 | Haloalkanes and HaloarenesSN1 and SN2 mechanisms | Class 12: Haloalkanes and Haloarenes | StatementsEasy | |
A easy statements question on sn1 and sn2 mechanisms, from Haloalkanes and Haloarenes. NCERT: Class 12: Haloalkanes and Haloarenes Statement I: in $\mathrm{S_N2}$ the nucleophile attacks from the back and the configuration is inverted (Walden inversion); a given stereoisomer of substrate gives one stereoisomer of product. True.
Statement II: in $\mathrm{S_N1}$ the planar carbocation is attacked from either face, so an optically active substrate gives a (largely) racemic product. True.
Answer: (A). | ||||
| 76 | Aldehydes, Ketones and Carboxylic AcidsAcidity of carboxylic acids | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctEasy | |
A easy single correct question on acidity of carboxylic acids, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Alkyl groups are electron-releasing (+I); they destabilise the carboxylate ion and weaken the acid. HCOOH has no alkyl group and is the strongest ($pK_a\approx3.75$); acetic acid ($4.76$), propanoic ($4.87$) and butanoic ($4.82$–$4.9$) follow, the +I effect growing with chain length.
Decreasing order: $\mathrm{HCOOH>CH_3COOH>CH_3CH_2COOH>CH_3CH_2CH_2COOH}$.
Answer: (B). | ||||
| 77 | Alcohols, Phenols and EthersReactions of phenols | Class 12: Alcohols, Phenols and Ethers | Matching listMedium | |
A medium matching list question on reactions of phenols, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers (A) Aniline with $\mathrm{NaNO_2+HCl}$ gives benzenediazonium chloride; warming with water replaces $\mathrm{N_2^+}$ by OH: phenol → (II).
(B) Phenol oxidised by $\mathrm{Na_2Cr_2O_7/H_2SO_4}$ gives p-benzoquinone → (IV).
(C) Phenol with $\mathrm{CHCl_3}$ and aqueous NaOH, then acid (Reimer–Tiemann reaction): salicylaldehyde (o-hydroxybenzaldehyde) → (I).
(D) Sodium phenoxide with $\mathrm{CO_2}$, then acid (Kolbe's reaction): salicylic acid (o-hydroxybenzoic acid) → (III).
A→II, B→IV, C→I, D→III.
Answer: (C). | ||||
| 78 | Alcohols, Phenols and EthersReactions of phenols | Class 12: Alcohols, Phenols and Ethers | Single correctEasy | |
A easy single correct question on reactions of phenols, from Alcohols, Phenols and Ethers. NCERT: Class 12: Alcohols, Phenols and Ethers Phenol ($pK_a\approx10$) is far more acidic than alcohols ($pK_a\approx16$–$18$) because the phenoxide ion is stabilised by resonance with the ring. It reacts readily with dilute NaOH: $\mathrm{C_6H_5OH+NaOH\to C_6H_5ONa+H_2O}$.
Benzyl alcohol, tert-butyl alcohol and ethanol are too weakly acidic to react with dilute aqueous NaOH.
Answer: (A). | ||||
| 79 | AminesClassification and identification of amines | Class 12: Amines | StatementsMedium | |
A medium statements question on classification and identification of amines, from Amines. NCERT: Class 12: Amines Sulphonyl chlorides (Hinsberg-type reagents) react only with amines that carry an N–H.
Statement I: aniline (primary) and diphenylamine (secondary) react, but triphenylamine (tertiary) has no N–H and does not. False.
Statement II: the sulphonamide from a primary amine ($\mathrm{ArSO_2NHC_6H_5}$) has an acidic N–H and dissolves in NaOH, but that from a secondary amine ($\mathrm{ArSO_2N(C_6H_5)_2}$) has none and is insoluble. False.
Answer: (B). | ||||
| 80 | Principles Related to Practical ChemistryDetection of elements and functional groups | No matching NCERT chapter | Matching listEasy | |
A easy matching list question on detection of elements and functional groups, from Principles Related to Practical Chemistry. NCERT: No matching NCERT chapter (A) Baeyer's test (cold dilute alkaline $\mathrm{KMnO_4}$ is decolourised): unsaturation → (IV).
(B) Ceric ammonium nitrate gives a red colour with alcohols: alcoholic –OH → (III).
(C) Phthalein dye test (fused with phthalic anhydride and conc. $\mathrm{H_2SO_4}$, then alkali): phenols → (I).
(D) Schiff's reagent turns pink/magenta with aldehydes → (II).
A→IV, B→III, C→I, D→II.
Answer: (A). | ||||
| 81 | SolutionsConcentration terms | Class 12: Solutions | NumericalEasy | |
A easy numerical question on concentration terms, from Solutions. NCERT: Class 12: Solutions Mass of ethanol $=1\times46=46$ g; mass of water $=9\times18=162$ g; total $=208$ g.
Mass percent of solute $=\frac{46}{208}\times100=22.1\%$, which rounds to 22.
Answer: 22. | ||||
| 82 | Structure of AtomDual nature and uncertainty principle | Class 11: Structure of Atom | NumericalEasy | |
A easy numerical question on dual nature and uncertainty principle, from Structure of Atom. NCERT: Class 11: Structure of Atom $\lambda=5800$ Å $=5800\times10^{-8}$ cm $=5.8\times10^{-5}$ cm.
Wavenumber $\bar\nu=\frac1\lambda=\frac{1}{5.8\times10^{-5}}=17241$ cm$^{-1}=1724.1\times10$ cm$^{-1}$.
$x=1724.1$, which rounds to 1724.
Answer: 1724. | ||||
| 83 | Chemical Bonding and Molecular StructureMolecular orbital theory | Class 11: Chemical Bonding and Molecular Structure | NumericalEasy | |
A easy numerical question on molecular orbital theory, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure Bond order $=\frac12(N_b-N_a)$ from MO theory:
$\mathrm{C_2}$ (12 e$^-$): $\frac12(8-4)=2$.
$\mathrm{O_2}$ (16 e$^-$): $\frac12(10-6)=2$.
$\mathrm{Be_2}$: 0. $\mathrm{Li_2}$: 1. $\mathrm{Ne_2}$: 0. $\mathrm{N_2}$: 3. $\mathrm{He_2}$: 0.
Bond order 2: $\mathrm{C_2}$ and $\mathrm{O_2}$, i.e. 2 molecules.
Answer: 2. | ||||
| 84 | ThermodynamicsFirst law and state functions | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on first law and state functions, from Thermodynamics. NCERT: Class 11: Thermodynamics $\mathrm{H_2O(l)\to H_2O(g)}$: $\Delta n_g=1$.
$\Delta U=\Delta H-\Delta n_gRT=40.79-\frac{1\times8.3\times373}{1000}=40.79-3.10=37.69$ kJ mol$^{-1}$.
This rounds to 38.
Answer: 38. | ||||
| 85 | SolutionsConcentration terms | Class 12: Solutions | NumericalEasy | |
A easy numerical question on concentration terms, from Solutions. NCERT: Class 12: Solutions 4.44 m: 4.44 mol urea in 1000 g water. Moles of water $=\frac{1000}{18}=55.56$.
Mole fraction of urea $=\frac{4.44}{4.44+55.56}=\frac{4.44}{60}=0.074=74\times10^{-3}$.
Answer: 74. | ||||
| 86 | Aldehydes, Ketones and Carboxylic AcidsAldol and Cannizzaro | Class 12: Aldehydes, Ketones and Carboxylic Acids | NumericalMedium | |
A medium numerical question on aldol and cannizzaro, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids Acetone has α-hydrogens on both methyl groups; each condenses with a benzaldehyde (Claisen–Schmidt, cross aldol condensation) and dehydrates on heating:
$2\,\mathrm{C_6H_5CHO}+\mathrm{CH_3COCH_3}\xrightarrow{\text{NaOH},\,\Delta}\mathrm{C_6H_5CH{=}CH{-}CO{-}CH{=}CHC_6H_5}$ (dibenzalacetone) $+2\mathrm{H_2O}$.
π bonds: 2 (C=C) + 1 (C=O) + 3 in each of the 2 benzene rings (6) = 9.
Answer: 9. | ||||
| 87 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | NumericalEasy | |
A easy numerical question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds $[\mathrm{Co(NH_3)_6}]^{3+}$: $\mathrm{Co^{3+}}$, $3d^6$; $\mathrm{NH_3}$ is strong field for $\mathrm{Co^{3+}}$, low spin $t_{2g}^6e_g^0$: 0 unpaired.
$[\mathrm{NiCl_4}]^{2-}$: $\mathrm{Ni^{2+}}$, $3d^8$; tetrahedral with weak-field $\mathrm{Cl^-}$ ($sp^3$): 2 unpaired.
Total $=0+2=2$.
Answer: 2. | ||||
| 88 | HydrocarbonsAromatic hydrocarbons | Class 11: Hydrocarbons | NumericalMedium | |
A medium numerical question on aromatic hydrocarbons, from Hydrocarbons. NCERT: Class 11: Hydrocarbons A compound is aromatic when it is cyclic, planar, fully conjugated (every ring atom $sp^2$) and has $(4n+2)$ π electrons. Taking the six structures in order:
1. The bicyclic triene: its left ring has a $\mathrm{CH_2}$ and the right ring two $\mathrm{CH_2}$ groups, so no ring is fully conjugated. Non-aromatic.
2. Two cyclopentene rings joined by a C=C: each ring has $sp^3$ $\mathrm{CH_2}$ carbons. Non-aromatic.
3. Cyclopentadienyl cation: planar, conjugated, 4 π electrons. Antiaromatic.
4. Cyclooctatetraene: 8 π electrons, tub-shaped. Non-aromatic.
5. Pyridine: planar, conjugated, 6 π electrons (the N lone pair lies in the plane). Aromatic.
6. Cycloheptatriene: one $sp^3$ $\mathrm{CH_2}$. Non-aromatic.
Only pyridine is aromatic: 1.
Answer: 1. | ||||
| 89 | Organic Chemistry: Some Basic Principles and TechniquesIsomerism | Class 11: Organic Chemistry – Some Basic Principles and Techniques | NumericalMedium | |
A medium numerical question on isomerism, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques 1. The Fischer projection of butane-2,3-diol has both H on the left and both OH on the right: it has a mirror plane, the meso form. Optically inactive.
2. Heptane-3,4,5-triol (drawn without stereochemistry): C3 and C5 are stereocentres, so the compound has chiral (optically active) stereoisomers. Counted as optically active.
3. Butan-1-ol: no stereocentre. Inactive.
4. 2-Chlorobutane: C2 bears H, Cl, $\mathrm{CH_3}$, $\mathrm{C_2H_5}$, a stereocentre. Optically active.
5. 1-Chlorobutane: inactive.
6. 1-Chloro-3-methylbutane: no stereocentre. Inactive.
Optically active: 2 (the triol and 2-chlorobutane).
Answer: 2. | ||||
| 90 | BiomoleculesProteins | Class 12: Biomolecules | NumericalEasy | |
A easy numerical question on proteins, from Biomolecules. NCERT: Class 12: Biomolecules Tyrosine is 2-amino-3-(4-hydroxyphenyl)propanoic acid, $\mathrm{HO{-}C_6H_4{-}CH_2{-}CH(NH_2){-}COOH}$, formula $\mathrm{C_9H_{11}NO_3}$.
Carbon atoms: 6 (ring) + 1 ($\mathrm{CH_2}$) + 1 (CH) + 1 (COOH) = 9.
Answer: 9. | ||||
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