JEE Main Full Mock Test: Practice Test 16
Question paper with answer key. Selection Pro
- $3:1$
- $1:1$
- $1:3$
- $3:4$
Answer: (A) $3:1$
$U/E=x^2/A^2=1/4$, so $K=3E/4$. Hence $K:U=3:1$.
- $\frac{2u^2}{3g}$
- $\frac{8u^2}{9g}$
- $\frac{4u^2}{9g}$
- $\frac{4\sqrt2u^2}{9g}$
Answer: (D) $\frac{4\sqrt2u^2}{9g}$
At the highest point $u\cos\theta=u/3$, so $\cos\theta=1/3$. Therefore $\sin2\theta=4\sqrt2/9$, giving $R=4\sqrt2u^2/(9g)$.
- $\sqrt{k/m}$
- $\sqrt{2k/m}$
- $\sqrt{m/2k}$
- $2\sqrt{k/m}$
Answer: (B) $\sqrt{2k/m}$
Both springs provide restoring force, so $F=-2kx$. Thus $m\ddot x+2kx=0$ and $\omega=\sqrt{2k/m}$.
- $\frac43ML^2$
- $\frac12ML^2$
- $ML^2$
- $\frac13ML^2$
Answer: (A) $\frac43ML^2$
For the rod $I=ML^2/3$ and for the end mass $I=ML^2$. Total $I=4ML^2/3$.
- $\sqrt{7gh/10}$
- $\sqrt{10gh/7}$
- $\sqrt{gh}$
- $\sqrt{2gh}$
Answer: (B) $\sqrt{10gh/7}$
Using $I=2MR^2/5$ and $v=\omega R$, energy conservation gives $Mgh=7Mv^2/10$. Hence $v=\sqrt{10gh/7}$.
- $1:27$
- $1:9$
- $3:1$
- $1:3$
Answer: (A) $1:27$
Kepler's law gives $T\propto r^{3/2}$, so $T_2/T_1=9^{3/2}=27$.
- $2CV^2$
- $\frac14CV^2$
- $CV^2$
- $\frac12CV^2$
Answer: (C) $CV^2$
Charge remains constant and capacitance becomes $C/2$. Since $U=Q^2/(2C)$, energy doubles to $CV^2$.
- $d/4$
- $d/2$
- $2d/3$
- $d/3$
Answer: (D) $d/3$
Equating field magnitudes: $q/x^2=4q/(d-x)^2$, hence $d-x=2x$ and $x=d/3$.
- $2R$
- $R/2$
- $4R$
- $R$
Answer: (C) $4R$
Constant volume gives $A' = A/2$. Since $R=\rho L/A$, doubling length and halving area gives $R'=4R$.
- $r/2,\ E^2/(2r)$
- $r,\ E^2/r$
- $r,\ E^2/(4r)$
- $2r,\ E^2/(2r)$
Answer: (C) $r,\ E^2/(4r)$
Maximum power transfer requires $R=r$. Then $P_{max}=E^2/(4r)$.
- $\frac{2\pi qB}{mv\cos\theta}$
- $\frac{2\pi mv\cos\theta}{qB}$
- $\frac{mv}{qB}$
- $\frac{2\pi mv\sin\theta}{qB}$
Answer: (B) $\frac{2\pi mv\cos\theta}{qB}$
Parallel speed is $v\cos\theta$ and cyclotron period is $2\pi m/(qB)$. Their product is the pitch.
- $B/(ARt)$
- $BAt/R$
- $BA/(Rt)$
- $BR/(At)$
Answer: (C) $BA/(Rt)$
Induced emf is $BA/t$, so $I=BA/(Rt)$.
- $\sqrt{R^2+(X_L+X_C)^2}$ and $\pi/2$
- $R$ and $\pi/2$
- $X_L-X_C$ and $0$
- $R$ and $0$
Answer: (D) $R$ and $0$
At resonance $X_L=X_C$, hence net reactance is zero, $Z=R$, and current is in phase with voltage.
- $110$ V
- $22$ V
- $2.2$ V
- $2200$ V
Answer: (B) $22$ V
$V_s/V_p=N_s/N_p=1/10$, hence $V_s=22$ V.
- $10$ cm
- $30$ cm
- $40$ cm
- $20$ cm
Answer: (B) $30$ cm
For a real image formed by a convex lens, magnification $m = -2$. Using $m = \frac{f}{f+u}$ with $f = 20\text{ cm}$, we get $-2 = \frac{20}{20+u} \implies -40 - 2u = 20 \implies u = -30\text{ cm}$. Hence, $|u| = 30\text{ cm}$.
- $I_0$
- $9I_0/16$
- $I_0/2$
- $I_0/4$
Answer: (B) $9I_0/16$
Original amplitude sum is $2a$; new sum is $3a/2$. Intensities are proportional to squared amplitudes, so ratio is $9/16$.
- $m_p/m_e$
- $1$
- $\sqrt{m_p/m_e}$
- $\sqrt{m_e/m_p}$
Answer: (C) $\sqrt{m_p/m_e}$
For the same potential, $p=\sqrt{2mqV}$ and $\lambda\propto1/\sqrt m$. Thus $\lambda_e/\lambda_p=\sqrt{m_p/m_e}$.
- $2$
- $1/\ln2$
- $\ln2$
- $1/2$
Answer: (B) $1/\ln2$
$\tau=1/\lambda$ and $T_{1/2}=\ln2/\lambda$, so the ratio is $1/\ln2$.
- $V/T=\text{constant}$
- $P/T=\text{constant}$
- $PV^\gamma=\text{constant}$
- $PV=\text{constant}$
Answer: (C) $PV^\gamma=\text{constant}$
The reversible adiabatic equation is $PV^\gamma=constant$.
- $3RT$
- $2RT$
- $\frac32RT$
- $RT$
Answer: (B) $2RT$
At constant pressure, $W=nR\Delta T=R(3T-T)=2RT$.
- $v$
- $4v$
- $2v$
- $v/4$
Answer: (B) $4v$
Continuity gives $Av=(A/4)v_2$, hence $v_2=4v$.
- $h/3$
- $h$
- $9h$
- $3h$
Answer: (D) $3h$
Capillary rise varies inversely with radius, so reducing radius by 3 increases rise by 3.
- $R/(\alpha t)$
- $\alpha Rt$
- $\alpha t^2/2$
- $\alpha t/R$
Answer: (B) $\alpha Rt$
Starting from rest, $\omega=\alpha t$. Therefore $v=R\omega=\alpha Rt$.
- $mgk$
- $mg/k$
- $g/k$
- $k/mg$
Answer: (B) $mg/k$
At equilibrium $kx=mg$, so $x=mg/k$.
- $1:4$
- $1:2$
- $2:1$
- $4:1$
Answer: (A) $1:4$
Equivalent resistances are $2R$ and $R/2$. Since $P=V^2/R$, the ratio is $1:4$.
- $pK_b$
- $pK_a$
- $7$
- $2pK_a$
Answer: (B) $pK_a$
Henderson-Hasselbalch gives $pH=pK_a+\log([A^-]/[HA])=pK_a$.
- $K_{sp}=4s^3$
- $K_{sp}=s^2$
- $K_{sp}=2s^2$
- $K_{sp}=s^3$
Answer: (A) $K_{sp}=4s^3$
$K_{sp}=(2s)^2(s)=4s^3$.
- $5$ min
- $20$ min
- $100$ min
- $10$ min
Answer: (D) $10$ min
$t_{1/2}=0.693/k=10$ min.
- 4
- 2
- 3
- 1
Answer: (C) 3
Overall order is $2+1=3$.
- $\Delta G^\circ<0$
- $K=0$
- $\Delta G^\circ=0$
- $\Delta G^\circ>0$
Answer: (A) $\Delta G^\circ<0$
$\Delta G^\circ=-nFE^\circ_{cell}$, so positive cell potential corresponds to negative standard Gibbs energy.
- $10$
- $1000$
- $100$
- $1$
Answer: (A) $10$
$E^\circ=(0.0591/n)\log K$. With $n=1$, $\log K=1$, hence $K=10$.
- $+7$
- $+5$
- $+6$
- $+3$
Answer: (C) $+6$
$2(+1)+2x+7(-2)=0$ gives $x=+6$.
- $NH_3$
- $BF_3$
- $H_2O$
- $CH_4$
Answer: (B) $BF_3$
$BF_3$ is trigonal planar with $120^\circ$ bond angles.
- Square planar
- See-saw
- Tetrahedral
- Trigonal bipyramidal
Answer: (A) Square planar
Six electron domains occur around Xe; two lone pairs occupy opposite positions, giving square-planar molecular geometry.
- $O_2^-$
- $O_2^{2-}$
- $O_2^+$
- $O_2$
Answer: (C) $O_2^+$
Removing one antibonding electron from O2 raises bond order from 2 to 2.5 in $O_2^+$.
- $Cl_3CCOOH$
- $ClCH_2COOH$
- $CH_3COOH$
- $Cl_2CHCOOH$
Answer: (A) $Cl_3CCOOH$
The -I effect of chlorine stabilizes the conjugate base; three Cl atoms give the strongest acid.
- 2-methylbutane
- 3-methyl-1-butene
- 2-methyl-2-butene
- 2-methyl-1-butene
Answer: (C) 2-methyl-2-butene
Acid-catalyzed dehydration predominantly gives the Zaitsev alkene, 2-methyl-2-butene.
- $K_2Cr_2O_7/H^+$
- PCC
- $KMnO_4$
- $LiAlH_4$
Answer: (B) PCC
PCC is a mild oxidant that stops primary alcohol oxidation at the aldehyde stage.
- p-bromophenol only
- No reaction
- Bromobenzene
- 2,4,6-tribromophenol
Answer: (D) 2,4,6-tribromophenol
The strongly activating -OH group directs bromination to both ortho and para positions, giving 2,4,6-tribromophenol.
- Cyclohexylamine has no lone pair
- Nitrogen in aniline is sp3
- Aniline is ionic
- Its lone pair is delocalized into the benzene ring
Answer: (D) Its lone pair is delocalized into the benzene ring
Resonance delocalization makes the nitrogen lone pair in aniline less available for protonation.
- 6
- 2
- 3
- 4
Answer: (D) 4
There are two stereogenic centers and no internal plane of symmetry, giving $2^2=4$ stereoisomers.
- $Br_2/CCl_4$
- Peroxide ($ROOR$)
- $Hg(OAc)_2$
- $H_2SO_4$
Answer: (B) Peroxide ($ROOR$)
Peroxide initiates a radical chain mechanism for anti-Markovnikov addition of HBr.
- Acetic acid only
- Acetone + formaldehyde
- Propanone only
- Methanol + CO
Answer: (B) Acetone + formaldehyde
Cleavage of $CH_2=C(CH_3)_2$ produces $HCHO$ and acetone.
- +4, 0
- +3, 1
- +2, 2
- +2, 0
Answer: (D) +2, 0
Fe is +2. It is d6 and CN- is strong-field, giving low-spin $t_{2g}^6$ with zero unpaired electrons.
- Trigonal planar
- Square planar
- Tetrahedral
- Octahedral
Answer: (B) Square planar
Pt(II) is d8 and commonly adopts square-planar geometry.
- NaCl
- CaO
- MgO
- KCl
Answer: (C) MgO
The 2+/2- ionic charges in MgO produce much stronger electrostatic attraction than the 1+/1- salts.
- Infinite
- Zero
- Positive
- Negative
Answer: (B) Zero
By definition, an ideal solution has $\Delta H_{mix}=0$.
- Increases vapour pressure
- Decreases molar mass
- Makes surface tension zero
- Decreases vapour pressure
Answer: (D) Decreases vapour pressure
The solute lowers solvent vapour pressure, requiring a higher temperature to reach external pressure.
- 4
- 5
- 3
- 2
Answer: (B) 5
Complete dissociation gives 2 Al3+ and 3 sulfate ions, i.e. 5 particles.
- Si
- Al
- Mg
- Na
Answer: (A) Si
Across Period 3, Si has the highest first ionization enthalpy among these four.
- $x=0$
- $y=0$
- $y=2$
- $x=2$
Answer: (A) $x=0$
The locus is the perpendicular bisector of $(2,0)$ and $(-2,0)$, hence $x=0$.
- 1
- 2
- 3
- 4
Answer: (B) 2
The four roots have modulus 2; the real ones are $2$ and $-2$.
- $-30$
- $-36$
- $20$
- $-20$
Answer: (C) $20$
Rewrite as $(1+x)^4(1-x^2)^5$. Expanding $(1+x)^4 = 1+4x+6x^2+4x^3+x^4$ and $(1-x^2)^5 = 1-5x^2+10x^4-\dots$. The $x^5$ term comes from $(4x)(10x^4) + (4x^3)(-5x^2) = 40x^5 - 20x^5 = 20x^5$. So the coefficient is $20$.
- $0$
- $1$
- $-1$
- $2$
Answer: (A) $0$
Discriminant $=4[(a+1)^2-(a^2+1)]=8a$. Equal roots require it to be zero, so $a=0$.
- $1$
- $-1$
- $0$
- Cannot be determined
Answer: (C) $0$
A determinant is zero whenever two rows or two columns are identical.
- Only 1
- Only 0
- $-1$ and $1$
- $0$ and $1$
Answer: (D) $0$ and $1$
For an eigenvalue $\lambda$, $\lambda^2=\lambda$, hence $\lambda=0$ or 1.
- $1/4$
- $1$
- $2$
- $1/2$
Answer: (D) $1/2$
For $x>0$, $1+x^2\ge2x$, so the ratio is at most $1/2$, attained at $x=1$.
- 2
- 3
- 1
- 0
Answer: (A) 2
$f'(x)=3(x^2-1)$ has critical points $x=\pm1$, and the derivative changes sign at both.
- $32/3$
- $8/3$
- $4$
- $16/3$
Answer: (D) $16/3$
$\int_0^4x^{1/2}dx=\frac23(4^{3/2})=16/3$.
- $1$
- $1/3$
- $2/3$
- $4/3$
Answer: (C) $2/3$
Using $\sin^3x=\sin x(1-\cos^2x)$ gives $\int_0^1(1-u^2)du=2/3$.
- $1/6$
- $1/3$
- $1/2$
- $2/3$
Answer: (B) $1/3$
Expansions give $\sin x=x-x^3/6+\cdots$ and $x\cos x=x-x^3/2+\cdots$, so the difference divided by $x^3$ tends to $1/3$.
- $y=x+C$
- $y=Cx$
- $y=Ce^x$
- $y=C/x$
Answer: (B) $y=Cx$
Separating gives $dy/y=dx/x$, hence $\ln|y|=\ln|x|+C$ and $y=Cx$.
- $\frac12(e^x+e^{-x})$
- $e^x-1$
- $xe^x$
- $\frac12(e^x-e^{-x})$
Answer: (D) $\frac12(e^x-e^{-x})$
The integrating factor is $e^x$. Thus $(ye^x)'=e^{2x}$, giving $y=\frac12(e^x-e^{-x})$ after applying $y(0)=0$.
- $4$
- $5$
- $3$
- $23/5$
Answer: (D) $23/5$
Distance is $|3(2)-4(-3)+5|/\sqrt{9+16}=23/5$.
- $x^2+y^2+6x+8y=0$
- $x^2+y^2=25$
- $x^2+y^2-6x-8y=0$
- $x^2+y^2-3x-4y=0$
Answer: (C) $x^2+y^2-6x-8y=0$
Radius squared is $25$. Expanding $(x-3)^2+(y-4)^2=25$ gives the stated equation.
- $3$
- $4$
- $\sqrt{34}$
- $2$
Answer: (B) $4$
Tangent length is $\sqrt{5^2-3^2}=4$.
- $9/25$
- $5/3$
- $3/5$
- $4/5$
Answer: (D) $4/5$
Here $a=5,b=3,c=4$, so $e=c/a=4/5$.
- $x+y=7$
- $4x+3y=25$
- $3x+4y=25$
- $3x-4y=25$
Answer: (C) $3x+4y=25$
The tangent formula is $xx_1+yy_1=r^2$, hence $3x+4y=25$.
- $5\sqrt2$
- $5$
- $7$
- $\sqrt{35}$
Answer: (A) $5\sqrt2$
$\vec a\times\vec b=(4,3,-5)$, whose magnitude is $\sqrt{16+9+25}=5\sqrt2$.
- $45^\circ$
- $30^\circ$
- $60^\circ$
- $90^\circ$
Answer: (C) $60^\circ$
Dot product is 1 and both magnitudes are $\sqrt2$, so $\cos\theta=1/2$ and $\theta=60^\circ$.
- $5/12$
- $7/18$
- $13/36$
- $1/2$
Answer: (A) $5/12$
Prime sums 2,3,5,7,11 have respectively 1,2,4,6,2 outcomes: 15 out of 36, giving $5/12$.
- $15$
- $10$
- $5$
- $20$
Answer: (B) $10$
The number of selections is $\binom53=10$.
- $5\%$
- $25\%$
- $20\%$
- $80\%$
Answer: (C) $20\%$
CV $=(4/20)\times100=20\%$.
- $25$
- $18$
- $225$
- $9$
Answer: (C) $225$
$Var(aX+b)=a^2Var(X)$, so $25\times9=225$.
- $9$
- $4$
- $6$
- $8$
Answer: (C) $6$
There are $2^3=8$ total functions and 2 constant non-onto functions, so $8-2=6$.