JEE Main Full Mock Test: Practice Test 13
Question paper with answer key. Selection Pro
- $\pi/3$
- $2\pi/3$
- $4\pi/3$
- $5\pi/3$
Answer: (A) $\pi/3$
$\cos\phi=1/2$ and $v=-A\omega\sin\phi<0$, so $\sin\phi>0$. Hence $\phi=\pi/3$.
- $\sqrt{3g/L}$
- $\sqrt{2g/L}$
- $\sqrt{6g/L}$
- $\sqrt{g/L}$
Answer: (A) $\sqrt{3g/L}$
$MgL/2=(1/2)(ML^2/3)\omega^2$, giving $\omega=\sqrt{3g/L}$.
- $\sqrt{k/m}$
- $\sqrt{2k/m}$
- $\sqrt{m/2k}$
- $2\sqrt{k/m}$
Answer: (B) $\sqrt{2k/m}$
The restoring force is $-2kx$, so $m\ddot x=-2kx$ and $\omega=\sqrt{2k/m}$.
- $30^\circ$
- $45^\circ$
- $60^\circ$
- $\cos^{-1}(1/3)$
Answer: (D) $\cos^{-1}(1/3)$
At the highest point the speed is the horizontal component $u\cos\theta$. Thus $\cos\theta=1/3$.
- $v/2$
- $v/4$
- $-v/2$
- $-v$
Answer: (C) $-v/2$
For an elastic one-dimensional collision, $v_1^\prime=(m-3m)/(m+3m)v=-v/2$.
- $1:3$
- $9:1$
- $3:1$
- $1:9$
Answer: (B) $9:1$
For circular orbit $K\propto1/r$, hence $K_R/K_{9R}=9$.
- $0$
- $kQ/R$
- $2kQ/R$
- $kQ/(2R)$
Answer: (B) $kQ/R$
Potential inside a charged conducting shell is constant and equal to its surface value $kQ/R$.
- $V$
- $5V/3$
- $4V/3$
- $2V$
Answer: (B) $5V/3$
Total initial charge is $CV+2C(2V)=5CV$ and total capacitance is $3C$, so $V_f=5V/3$.
- $R=r/2$
- $R=r$
- $R=2r$
- $R\to\infty$
Answer: (B) $R=r$
$P=E^2R/(R+r)^2$ is maximum at $R=r$.
- $B/4$
- $B/2$
- $B$
- $2B$
Answer: (C) $B$
$B=\mu_0I/(2\pi r)$. Doubling both $I$ and $r$ leaves $I/r$ unchanged.
- $\pi/\sqrt3$
- $2\pi/\sqrt3$
- $\sqrt3\pi$
- $2\pi\sqrt3$
Answer: (B) $2\pi/\sqrt3$
$p=v\cos60^\circ(2\pi m/qB)$ and $r=v\sin60^\circ m/qB$, giving $p/r=2\pi/\sqrt3$.
- $N\Phi/R$
- $N\Phi/(Rt)$
- $N\Phi t/R$
- $\Phi/(NR)$
Answer: (A) $N\Phi/R$
$q=(1/R)\int|\mathcal E|dt=N|\Delta\Phi|/R=N\Phi/R$.
- $0$
- $50$ V
- $100$ V
- $200$ V
Answer: (C) $100$ V
At resonance $X_L=X_C$, so $V_L$ and $V_C$ are equal in magnitude and opposite in phase.
- $1/2$
- $1/3$
- $2$
- $3$
Answer: (A) $1/2$
For $u=3f$, lens formula gives $v=3f/2$, hence $|m|=v/u=1/2$.
- $I$
- $3I$
- $5I$
- $9I$
Answer: (A) $I$
$I_R=I+4I+2\sqrt{4I^2}\cos\pi=I$.
- $2$ V
- $3$ V
- $4$ V
- $5$ V
Answer: (B) $3$ V
Photon energy is $1240/248=5$ eV. Thus $K_{max}=3$ eV and stopping potential is $3$ V.
- $\ln2/\lambda$
- $2\ln2/\lambda$
- $3\ln2/\lambda$
- $4\ln2/\lambda$
Answer: (D) $4\ln2/\lambda$
$A=A_0e^{-\lambda t}$; $1/16=e^{-4\ln2}$, so $t=4\ln2/\lambda$.
- $T/2$
- $T$
- $2T$
- $4T$
Answer: (C) $2T$
At constant pressure, $V/T$ is constant, so doubling $V$ doubles $T$.
- $PV=\text{constant}$
- $PV^\gamma=\text{constant}$
- $P/T=\text{constant}$
- $PT=\text{constant}$
Answer: (B) $PV^\gamma=\text{constant}$
For a reversible adiabatic process, $PV^\gamma$ is constant.
- $h/3$
- $h$
- $3h$
- $9h$
Answer: (C) $3h$
$h\propto1/r$, so reducing radius to $r/3$ makes the rise $3h$.
- $\Delta L/2$
- $\Delta L$
- $2\Delta L$
- $4\Delta L$
Answer: (B) $\Delta L$
$\Delta L^\prime=F(2L)/(2AY)=\Delta L$.
- $1:1$
- $1:2$
- $2:1$
- $1:4$
Answer: (B) $1:2$
Upward time is $u/g$ and total time is $2u/g$.
- $\tan\theta=\mu$
- $\tan\theta=1/\mu$
- $\sin\theta=\mu$
- $\cos\theta=\mu$
Answer: (A) $\tan\theta=\mu$
$mg\sin\theta=\mu mg\cos\theta$, hence $\tan\theta=\mu$.
- $\sqrt{2gh}$
- $\sqrt{10gh/7}$
- $\sqrt{7gh/10}$
- $\sqrt{gh}$
Answer: (B) $\sqrt{10gh/7}$
$mgh=\frac12mv^2+\frac12(2mR^2/5)(v^2/R^2)=7mv^2/10$, so $v=\sqrt{10gh/7}$.
- $T/4$
- $T/2$
- $2T$
- $4T$
Answer: (B) $T/2$
$T\propto1/\sqrt g$; with $g^\prime=4g$, $T^\prime=T/2$.
- $pK_a$
- $2pK_a$
- $pK_a/2$
- $14-pK_a$
Answer: (A) $pK_a$
$pH=pK_a+\log([A^-]/[HA])=pK_a$.
- $0.001$
- $0.01$
- $0.1$
- $0.1\%$
Answer: (B) $0.01$
$\alpha\approx\sqrt{K_a/C}=\sqrt{10^{-4}}=0.01$.
- $K_p=K_cRT$
- $K_p=K_c/(RT)$
- $K_p=K_c(RT)^2$
- $K_p=K_c$
Answer: (A) $K_p=K_cRT$
$K_p=K_c(RT)^{\Delta n}$ and $\Delta n=1$.
- $2t_{1/2}$
- $4t_{1/2}$
- $6.64t_{1/2}$
- $10t_{1/2}$
Answer: (C) $6.64t_{1/2}$
Remaining fraction is $0.01$. Thus $t/t_{1/2}=\log_2(100)=6.64$.
- $15$ min
- $20$ min
- $30$ min
- $60$ min
Answer: (C) $30$ min
$t_{1/2}=0.693/k\approx30$ min.
- Backward
- Forward
- Not at all
- Randomly
Answer: (B) Forward
The system must form more products until $Q_c$ reaches $K_c$.
- $Fe^{2+}$
- $Fe^{3+}$
- $Co^{2+}$
- $Ni^{2+}$
Answer: (B) $Fe^{3+}$
$Fe^{3+}$ is $3d^5$ and has five unpaired electrons.
- $\sqrt3$ BM
- $2\sqrt2$ BM
- $\sqrt{15}$ BM
- $3$ BM
Answer: (C) $\sqrt{15}$ BM
$\mu=\sqrt{n(n+2)}$ BM; for $n=3$, $\mu=\sqrt{15}$ BM.
- Tetrahedral, $sp^3$
- Square planar, $dsp^2$
- Octahedral, $d^2sp^3$
- Linear, $sp$
Answer: (B) Square planar, $dsp^2$
$CN^-$ is strong field; $Ni^{2+}$ forms a square-planar $dsp^2$ complex.
- $\Delta G^\circ>0$
- $\Delta G^\circ=0$
- $\Delta G^\circ<0$
- $K<1$
Answer: (C) $\Delta G^\circ<0$
$\Delta G^\circ=-nFE^\circ_{cell}$, so positive $E^\circ$ means negative $\Delta G^\circ$.
- Increase
- Decrease
- Unchanged
- Infinite
Answer: (B) Decrease
Nernst equation: $E=E^\circ-(RT/nF)\ln Q$; increasing $Q$ lowers $E$.
- $+3$
- $+4$
- $+6$
- $+7$
Answer: (C) $+6$
$2x-14=-2$ gives $x=+6$.
- $H_2O$
- $NH_3$
- $CO_2$
- $SO_2$
Answer: (C) $CO_2$
Linear symmetric $CO_2$ has equal and opposite bond dipoles.
- $1.5$
- $2$
- $2.5$
- $3$
Answer: (C) $2.5$
Removing one electron from antibonding $\pi^*$ raises the $O_2$ bond order from 2 to 2.5.
- LiF
- NaF
- KF
- RbF
Answer: (A) LiF
With the same charges, smaller ionic size gives stronger electrostatic attraction; $LiF$ is highest.
- 1-bromopropane
- 2-bromopropane
- 1,2-dibromopropane
- Propane
Answer: (A) 1-bromopropane
Peroxide effect gives anti-Markovnikov radical addition of HBr.
- Excess $KMnO_4$
- PCC
- Excess $K_2Cr_2O_7/H^+$
- $O_3$
Answer: (B) PCC
PCC oxidises a primary alcohol to an aldehyde under controlled conditions.
- Phenol has lower mass
- Phenoxide is resonance-stabilised
- Ethanol is aromatic
- Phenol has no O-H bond
Answer: (B) Phenoxide is resonance-stabilised
Resonance delocalisation stabilises the phenoxide ion.
- Acetone
- Propanone
- Ethanal
- Diethyl ether
Answer: (C) Ethanal
Aldehydes reduce Tollens reagent; ethanal is an aldehyde.
- It has no nitrogen
- Its N lone pair is delocalised into the benzene ring
- It is aromatic
- It has no lone pair
Answer: (B) Its N lone pair is delocalised into the benzene ring
Resonance makes the nitrogen lone pair in aniline less available for protonation.
- $1$
- $2$
- $3$
- $4$
Answer: (B) $2$
There are cis and trans forms, so two geometrical isomers.
- Polyethene
- PVC
- Nylon-6,6
- Polystyrene
Answer: (C) Nylon-6,6
Nylon-6,6 forms by condensation of a diamine and a dicarboxylic acid.
- $0$
- $1$
- Always greater than 1
- Always less than 1
Answer: (B) $1$
Ideal solutions obey Raoult law and have activity coefficients equal to one.
- Elevation of boiling point
- Depression of freezing point
- Osmotic pressure
- Relative lowering of vapour pressure
Answer: (C) Osmotic pressure
Osmotic pressure is especially sensitive and measurable for dilute macromolecular solutions.
- $+1$ kJ mol$^{-1}$
- $0$
- $-1$ kJ mol$^{-1}$
- Always positive
Answer: (B) $0$
By definition it is zero for an element in its standard state.
- Real axis
- Imaginary axis
- Unit circle
- $y=x$
Answer: (B) Imaginary axis
Writing $z=x+iy$ and squaring gives $(x-1)^2+y^2=(x+1)^2+y^2$, hence $x=0$.
- $1$
- $2$
- $3$
- $4$
Answer: (A) $1$
The roots are $2,2i,-2,-2i$; only $2$ has positive real part.
- $18$
- $20$
- $22$
- $24$
Answer: (C) $22$
For $i+2j=5$, $(i,j)=(1,2),(3,1)$. Coefficient $=\binom31\binom42+\binom33\binom41=18+4=22$.
- $b=0$
- $a=0$
- $a=b$
- $a=-b$
Answer: (A) $b=0$
Discriminant is $4b^2$. Equal roots require $b=0$.
- $a(r^n-1)/(r-1)$
- $a(1-r^n)/(1-r)$
- Both A and B
- Neither
Answer: (C) Both A and B
The first two expressions are algebraically identical.
- $-3$
- $1/3$
- $3$
- $9$
Answer: (B) $1/3$
$|A^{-1}|=1/|A|=1/3$.
- Perpendicular
- Parallel or antiparallel
- Equal only
- Always identical
Answer: (B) Parallel or antiparallel
$ab\sin\theta=0$ with non-zero magnitudes gives $\theta=0$ or $\pi$.
- $30^\circ$
- $45^\circ$
- $60^\circ$
- $90^\circ$
Answer: (C) $60^\circ$
Dot product is 1 and magnitudes are $\sqrt2$ each, so $\cos\theta=1/2$.
- $y=-2$
- $y=2x-4$
- $y=-2x$
- $y=2x$
Answer: (A) $y=-2$
At $(1,-2)$ the derivative is $3(1)^2-3=0$, hence the tangent is $y=-2$.
- $256/27$
- $128/27$
- $64/9$
- $32/9$
Answer: (A) $256/27$
Using f^\prime=x(8-3x), the interior maximum is at x=8/3, giving f=256/27.
- $1/3$
- $1/2$
- $1$
- $2/3$
Answer: (A) $1/3$
Using expansions, numerator $=x^3/3+O(x^5)$, so the limit is $1/3$.
- $x^x[(\ln x+1)^2+1/x]$
- $x^x(\ln x+1)$
- $x^{x-1}$
- $x^x(\ln x)^2$
Answer: (A) $x^x[(\ln x+1)^2+1/x]$
$y^\prime=x^x(\ln x+1)$; differentiating gives $y^{\prime\prime}=x^x[(\ln x+1)^2+1/x]$.
- $\frac1a\tan^{-1}(x/a)+C$
- $a\tan^{-1}(x/a)+C$
- $\ln(x^2+a^2)+C$
- $1/(x^2+a^2)+C$
Answer: (A) $\frac1a\tan^{-1}(x/a)+C$
This is the standard integral $\frac1a\tan^{-1}(x/a)+C$.
- $1/3$
- $2/3$
- $1$
- $4/3$
Answer: (B) $2/3$
Write $\sin^3x=\sin x(1-\cos^2x)$ and put $u=\cos x$; the result is $2/3$.
- $2/3$
- $4/3$
- $8/3$
- $4$
Answer: (B) $4/3$
Intersections are 0 and 2. Area $=\int_0^2(2x-x^2)dx=4/3$.
- $1/5$
- $1$
- $2$
- $3$
Answer: (A) $1/5$
Distance $=|3+8-12|/5=1/5$.
- $(x-3)^2+(y-4)^2=8$
- $(x-3)^2+(y-4)^2=4$
- $(x-2)^2+(y-3)^2=8$
- $(x-3)^2+(y-4)^2=16$
Answer: (A) $(x-3)^2+(y-4)^2=8$
Centre is $(3,4)$ and radius squared is $[(4)^2+(4)^2]/4=8$.
- $3/5$
- $4/5$
- $5/3$
- $2/5$
Answer: (B) $4/5$
$e=\sqrt{1-b^2/a^2}=\sqrt{1-9/25}=4/5$.
- $y^2=8x$
- $y^2=4x$
- $x^2=8y$
- $y^2=16x$
Answer: (A) $y^2=8x$
Its vertex is origin and focal length is 2. Thus $y^2=4ax=8x$.
- $x+y=1$
- $xy=1$
- $x+y+xy=1$
- $x-y=1$
Answer: (C) $x+y+xy=1$
Taking tangent gives $(x+y)/(1-xy)=1$, hence $x+y+xy=1$.
- $3/5$
- $1/2$
- $7/10$
- $2/5$
Answer: (A) $3/5$
By inclusion-exclusion: $50+20-10=60$ favourable integers, so probability $=3/5$.
- $5/12$
- $1/2$
- $7/18$
- $1/3$
Answer: (A) $5/12$
Prime sums 2,3,5,7,11 have $1+2+4+6+2=15$ outcomes, so $15/36=5/12$.
- $(n^2-1)/12$
- $(n^2+1)/12$
- $n(n+1)/12$
- $n^2/12$
Answer: (A) $(n^2-1)/12$
The standard result for $1,2,\ldots,n$ is $(n^2-1)/12$.
- $12$
- $13$
- $14$
- $15$
Answer: (C) $14$
Old sum is 60; new sum is $60-8+18=70$, so new mean is 14.
- $20$
- $30$
- $60$
- $120$
Answer: (B) $30$
Two Ls and two Es repeat: $5!/(2!2!)=30$.