JEE Main Full Mock Test: Practice Test 06
Question paper with answer key. Selection Pro
- $1$
- $2$
- $4$
- $1/2$
Answer: (C) $4$
Using $H=u^2\sin^2\theta/(2g)$ and $R=u^2\sin2\theta/g$, the condition $H=R/4$ gives $\tan\theta=2$.
- $\sqrt{k/m}$
- $\sqrt{2k/m}$
- $2\sqrt{k/m}$
- $\sqrt{m/2k}$
Answer: (B) $\sqrt{2k/m}$
Both springs provide restoring forces in the same direction, so $F=-2kx$. Hence $\omega=\sqrt{2k/m}$.
- $\sqrt{g/L}$
- $\sqrt{2g/L}$
- $\sqrt{3g/L}$
- $\sqrt{6g/L}$
Answer: (C) $\sqrt{3g/L}$
The centre of mass falls by $L/2$. Thus $MgL/2=(1/2)(ML^2/3)\omega^2$, giving $\omega=\sqrt{3g/L}$.
- $\sqrt{k/m}\left(1+\frac{3\alpha A^2}{2k}\right)$
- $\sqrt{k/m}\left(1-\frac{3\alpha A^2}{2k}\right)$
- $\sqrt{k/m}\left(1+\frac{3\alpha A^2}{k}\right)$
- $\sqrt{k/m}$
Answer: (A) $\sqrt{k/m}\left(1+\frac{3\alpha A^2}{2k}\right)$
The equation is $\ddot x+(k/m)x+(4\alpha/m)x^3=0$. The first-order Duffing correction gives $\omega\approx\sqrt{k/m}[1+3\alpha A^2/(2k)]$.
- $1/4$
- $1/2$
- $2$
- $4$
Answer: (A) $1/4$
For a circular orbit $E=-GMm/(2r)$. Hence $E_f/E_i=R/(4R)=1/4$.
- $q/\varepsilon_0$
- $q/(2\varepsilon_0)$
- $q/(6\varepsilon_0)$
- $q/(8\varepsilon_0)$
Answer: (C) $q/(6\varepsilon_0)$
Gauss's law gives total flux $q/\varepsilon_0$. By symmetry it is equally divided among six faces.
- $V/2$
- $3V/4$
- $3V/2$
- $V$
Answer: (B) $3V/4$
Initially, the total charge is $(C+2C)V=3CV$. After the dielectric is inserted, the capacitance $C$ becomes $2C$, so the total capacitance is $4C$. Since the system is isolated, charge remains constant. Therefore $V_f=3CV/(4C)=3V/4$.
- $R=0$
- $R=r$
- $R=2r$
- $R\to\infty$
Answer: (B) $R=r$
The external power is $P=E^2R/(R+r)^2$. Differentiation gives its maximum at $R=r$.
- $1/3$
- $1/2$
- $2/3$
- $1/4$
Answer: (A) $1/3$
Current divides inversely to resistance. Thus $I_{2R}:I_R=R:2R=1:2$, so $I_{2R}=I_{total}/3$.
- $\frac{\sin\theta}{2\pi\cos\theta}$
- $\frac{\cos\theta}{2\pi\sin\theta}$
- $\frac{\sin\theta}{\cos\theta}$
- $2\pi\tan\theta$
Answer: (A) $\frac{\sin\theta}{2\pi\cos\theta}$
The radius of the helical path is $r=mv\sin\theta/(qB)$, while the pitch is $p=v\cos\theta(2\pi m/qB)$. Hence $r/p=\tan\theta/(2\pi)=\sin\theta/(2\pi\cos\theta)$, which is option 1.
- $NBA$
- $NBA\omega$
- $NBA/\omega$
- $B/(NA\omega)$
Answer: (B) $NBA\omega$
Magnetic flux is $NBA\cos\omega t$, so the maximum magnitude of $-d\Phi/dt$ is $NBA\omega$.
- $I/2$
- $I$
- $2I$
- $I/\sqrt2$
Answer: (A) $I/2$
At resonance impedance equals $R$. Doubling $R$ halves the current.
- $+5/3$ D
- $+1/60$ D
- $-1/60$ D
- $+1/5$ D
Answer: (A) $+5/3$ D
The powers are $+5$ D and $-10/3$ D. Hence $P=5-10/3=5/3$ D.
- $\beta/2$
- $\beta/4$
- $2\beta$
- $4\beta$
Answer: (B) $\beta/4$
Since $\beta=\lambda D/d$, replacing $\lambda$ by $\lambda/2$ and $d$ by $2d$ makes $\beta'=\beta/4$.
- $3$ V
- $5$ V
- $8$ V
- $11$ V
Answer: (B) $5$ V
Einstein's equation gives $K_{max}=8-3=5$ eV. Therefore stopping potential is $5$ V.
- $1$
- $\sqrt{m_p/m_e}$
- $\sqrt{m_e/m_p}$
- $m_p/m_e$
Answer: (B) $\sqrt{m_p/m_e}$
For the same accelerating potential, $\lambda=h/\sqrt{2mqV}$, so $\lambda_e/\lambda_p=\sqrt{m_p/m_e}$.
- $1/2$
- $1/4$
- $1/8$
- $1/16$
Answer: (D) $1/16$
Twenty-four hours correspond to four half-lives, so the fraction left is $(1/2)^4=1/16$.
- $PV=constant$
- $PV^\gamma=constant$
- $P/V^\gamma=constant$
- $P+V^\gamma=constant$
Answer: (B) $PV^\gamma=constant$
For a reversible adiabatic process of an ideal gas, $PV^\gamma=constant$.
- $RT\ln2$
- $RT\ln4$
- $4RT\ln4$
- $3RT$
Answer: (B) $RT\ln4$
For one mole in an isothermal process, $W=RT\ln(V_2/V_1)=RT\ln4$.
- $g(\sin\theta-\mu\cos\theta)$
- $g(\sin\theta+\mu\cos\theta)$
- $g(\cos\theta+\mu\sin\theta)$
- $g\sin\theta$
Answer: (B) $g(\sin\theta+\mu\cos\theta)$
Both gravity's component and friction act down the incline, so the magnitude of acceleration is $g(\sin\theta+\mu\cos\theta)$.
- $3:4$
- $4:3$
- $1:2$
- $2:1$
Answer: (A) $3:4$
For the cylinder $I=MR^2/2$, giving $K=3Mv^2/4$. For the ring $I=MR^2$, giving $K=Mv^2$. Ratio is $3:4$.
- $1:1$
- $2:1$
- $4:1$
- $8:1$
Answer: (B) $2:1$
The new radius is $R/2$. Total area becomes $8\times4\pi(R/2)^2=2(4\pi R^2)$, so surface energy doubles.
- $1/2$
- $2$
- $4$
- $8$
Answer: (B) $2$
Doubling the radius gives four times the area, while doubling the length doubles the denominator in $k=YA/L$. Hence the new spring constant is $2k$.
- $T/2$
- $T\sqrt2$
- $2T$
- $4T$
Answer: (C) $2T$
At height $h=R$, $g'=GM/(2R)^2=g/4$. Since $T\propto1/\sqrt g$, the new period is $2T$.
- $pE\cos\theta$
- $pE\sin\theta$
- $p/E$
- $E/p$
Answer: (B) $pE\sin\theta$
The torque on a dipole is $\tau=pE\sin\theta$.
- 40 min
- 60 min
- 80 min
- 100 min
Answer: (C) 80 min
6.25% remains, which is $1/16=(1/2)^4$. Therefore four half-lives are required: $80$ min.
- $0.20$ M
- $0.50$ M
- $0.80$ M
- $1.20$ M
Answer: (C) $0.80$ M
Let $B=x$. Then $A=1-x$ and $K=x/(1-x)=4$. Hence $x=0.8$ M.
- $2$
- $4$
- $5$
- $8$
Answer: (B) $4$
For a weak acid, $[H^+]\approx\sqrt{K_aC}=\sqrt{10^{-8}}=10^{-4}$ M, so pH $\approx4$.
- $2.38$
- $4.76$
- $7.00$
- $9.52$
Answer: (B) $4.76$
Henderson-Hasselbalch equation gives $pH=pK_a+\log([salt]/[acid])=pK_a$.
- $-106.1$ kJ mol$^{-1}$
- $-212.3$ kJ mol$^{-1}$
- $+212.3$ kJ mol$^{-1}$
- $+106.1$ kJ mol$^{-1}$
Answer: (B) $-212.3$ kJ mol$^{-1}$
Using $\Delta G^\circ=-nFE^\circ$, $=-2\times96485\times1.10\approx-212.3$ kJ mol$^{-1}$.
- increases
- decreases
- remains unchanged
- becomes zero
Answer: (B) decreases
The reaction quotient increases, so from the Nernst equation $E=E^\circ-(0.0591/2)\log Q$, the cell potential decreases.
- 2-methyl-1-butene
- 2-methyl-2-butene
- 1-pentene
- 2-pentanol
Answer: (B) 2-methyl-2-butene
Dehydration proceeds mainly by an E1 mechanism and gives the more substituted alkene, 2-methyl-2-butene.
- CH3Cl
- CH3CH2Cl
- (CH3)2CHCl
- (CH3)3CCl
Answer: (D) (CH3)3CCl
SN1 rate depends strongly on carbocation stability. The tertiary carbocation is the most stable.
- 1-bromopropane
- 2-bromopropane
- 1,2-dibromopropane
- propanol
Answer: (A) 1-bromopropane
In the presence of peroxide, HBr undergoes radical anti-Markovnikov addition, producing 1-bromopropane.
- Cyclobutadiene
- Cyclopentadienyl anion
- Cyclopentadienyl cation
- Cycloheptatrienyl anion
Answer: (B) Cyclopentadienyl anion
The cyclopentadienyl anion has 6 $\pi$ electrons and is planar/conjugated, satisfying the $4n+2$ rule with $n=1$.
- 2
- 3
- 4
- 6
Answer: (B) 3
It has two stereocentres and a meso form, giving two enantiomers plus one meso form: three stereoisomers.
- Phenol
- p-Cresol
- p-Nitrophenol
- Cyclohexanol
Answer: (C) p-Nitrophenol
The nitro group withdraws electron density and stabilizes the phenoxide ion, making p-nitrophenol the strongest acid here.
- m-Nitroanisole
- p-Nitroanisole
- Nitrobenzene
- Benzoic acid
Answer: (B) p-Nitroanisole
The methoxy group is strongly activating and ortho/para directing; the para product is favored because of lower steric hindrance.
- Nitrobenzene
- Benzenediazonium chloride
- Phenol
- Benzene
Answer: (B) Benzenediazonium chloride
Primary aromatic amines undergo diazotisation with nitrous acid in cold acidic medium.
- $KMnO_4$
- $LiAlH_4$
- $K_2Cr_2O_7$
- $Br_2/H_2O$
Answer: (B) $LiAlH_4$
$LiAlH_4$ reduces aldehydes to primary alcohols.
- CH3COOH
- ClCH2COOH
- Cl2CHCOOH
- Cl3CCOOH
Answer: (D) Cl3CCOOH
The electron-withdrawing $-I$ effect of chlorine stabilizes the carboxylate ion. More chlorine atoms increase acidity.
- 4
- 5
- 6
- 7
Answer: (C) 6
Four ammonia ligands and two chloride ligands are directly coordinated, so the coordination number is 6.
- Tetrahedral
- Square planar
- Octahedral
- Trigonal planar
Answer: (B) Square planar
$CN^-$ is a strong-field ligand. For $Ni^{2+}$, pairing leads to a square-planar $dsp^2$ complex.
- 0
- 1
- $RT$
- $P/V$
Answer: (B) 1
The ideal-gas equation is $PV=nRT$, so $Z=1$.
- positive
- negative
- zero
- infinite
Answer: (C) zero
In an ideal solution, intermolecular interactions are effectively unchanged on mixing, so $\Delta H_{mix}=0$.
- 1
- 3
- 5
- 0
Answer: (C) 5
$Fe$ is $[Ar]3d^64s^2$; $Fe^{3+}$ is $[Ar]3d^5$, with five unpaired electrons by Hund's rule.
- NH3
- H2O
- BF3
- CH4
Answer: (C) BF3
$BF_3$ is trigonal planar with a $120^\circ$ bond angle, larger than the other listed molecules.
- Reactants
- Products
- Neither side
- Both equally
Answer: (B) Products
There are four gaseous moles on the reactant side and two on the product side, so increased pressure favors products.
- +3
- +4
- +6
- +7
Answer: (C) +6
Let chromium be $x$: $2(+1)+2x+7(-2)=0$, giving $x=+6$.
- $I^-$
- $F^-$
- $H_2O$
- $CN^-$
Answer: (D) $CN^-$
$CN^-$ lies toward the strong-field end of the spectrochemical series and produces large crystal-field splitting.
- $a>1$
- $a<-1$
- $|a|<1$
- $a\ge1$
Answer: (A) $a>1$
The roots are $a\pm1$. Both are positive iff $a-1>0$, hence $a>1$.
- 0
- 1
- 2
- 4
Answer: (C) 2
For $x\ge1$, the left side is $2x=4$, giving $x=2$. For $x\le-1$, it is $-2x=4$, giving $x=-2$. Thus there are two solutions.
- $x=0$
- $y=0$
- $x=2$
- $y=2$
Answer: (A) $x=0$
Squaring gives $(x-2)^2+y^2=(x+2)^2+y^2$, hence $x=0$.
- $1/3$
- $1/2$
- $2/3$
- $1$
Answer: (A) $1/3$
Using $\tan x=x+x^3/3+O(x^5)$, the limit is $1/3$.
- $e^e$
- $2e^e$
- $e^{e+1}$
- $e$
Answer: (B) $2e^e$
Since $f'(x)=x^x(\ln x+1)$, at $x=e$ it is $e^e(1+1)=2e^e$.
- 4
- 8
- 16
- 64
Answer: (C) 16
Since $x(4-x)=4-(x-2)^2\le4$, its square is at most $16$, attained at $x=2$.
- $-\pi\ln2$
- $-\frac{\pi}{2}\ln2$
- $0$
- $\frac{\pi}{2}\ln2$
Answer: (B) $-\frac{\pi}{2}\ln2$
The standard integral is $\int_0^{\pi/2}\ln(\sin x)dx=-\frac{\pi}{2}\ln2$.
- 2
- 4
- 8
- 16
Answer: (B) 4
For an $n\times n$ matrix, $|\operatorname{adj}A|=|A|^{n-1}$. Hence it is $2^2=4$.
- 3
- 4
- 5
- 6
Answer: (C) 5
$\sin x(1+2\cos x)=0$. Thus $x=0,\pi,2\pi$ and $\cos x=-1/2$ gives $2\pi/3,4\pi/3$, totaling 5.
- $0$
- $1$
- $2$
- $5$
Answer: (D) $5$
$\sin x(1+2\cos x)=0$. The solutions in $[0,2\pi]$ are $0,\pi,2\pi,2\pi/3,4\pi/3$, giving 5 solutions.
- $-14$
- $-8$
- $8$
- $14$
Answer: (A) $-14$
The coefficient of $x^5$ in $(1+x)^8(1-x)^4$ is $\sum_{j=0}^4(-1)^j\binom4j\binom8{5-j}=-14$.
- $x^2+y^2-4x-6y=0$
- $x^2+y^2+4x+6y=0$
- $x^2+y^2-6x-4y=0$
- $x^2+y^2+6x-4y=0$
Answer: (A) $x^2+y^2-4x-6y=0$
For $x^2+y^2+Dx+Ey=0$, substituting $(4,0)$ gives $D=-4$ and substituting $(0,6)$ gives $E=-6$.
- $3/4$
- $\sqrt7/4$
- $1/4$
- $\sqrt7/3$
Answer: (B) $\sqrt7/4$
The ellipse is $x^2/16+y^2/9=1$, so $a=4,b=3$. Hence $e=\sqrt{1-b^2/a^2}=\sqrt7/4$.
- 10
- 11
- 12
- 15
Answer: (B) 11
The minimum power is 0 and maximum is 10, and every integer power from 0 to 10 occurs. Hence there are 11 terms.
- $1/2$
- $1$
- $2$
- $4$
Answer: (B) $1$
This is a geometric series with first term $1/2$ and ratio $1/2$, so sum $=(1/2)/(1-1/2)=1$.
- 9
- 18
- 21
- 27
Answer: (B) 18
$\alpha+\beta=3$, $\alpha\beta=1$. Thus $\alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)=27-9=18$.
- $33/100$
- $1/3$
- $1/2$
- $1/6$
Answer: (A) $33/100$
Multiples of 2 or 3 among 1 to 100 are $50+33-16=67$. Hence 33 integers are divisible by neither, so probability is $33/100$.
- $y=x+C$
- $y=Cx$
- $y=C/x$
- $y=Ce^x$
Answer: (B) $y=Cx$
Separating variables gives $dy/y=dx/x$. Hence $\ln|y|=\ln|x|+C$, so $y=Cx$.
- $1/3$
- $1/6$
- $1/2$
- $2/3$
Answer: (B) $1/6$
The curves intersect at $x=0,1$. The area is $\int_0^1(x-x^2)dx=1/2-1/3=1/6$.
- $5\sqrt3$
- $\sqrt{35}$
- $5$
- $7$
Answer: (A) $5\sqrt3$
$\vec a\times\vec b=(5,5,-5)$, so its magnitude is $\sqrt{25+25+25}=5\sqrt3$.
- $1/\sqrt2$
- $1/\sqrt3$
- $\sqrt2$
- $1$
Answer: (A) $1/\sqrt2$
Using $d=| (\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)|/|\vec b_1\times\vec b_2|$ gives $1/\sqrt2$.
- 29
- 30
- 15
- 31
Answer: (A) 29
Mean is $(n+1)/2=15$, so $n+1=30$ and $n=29$.
- $2$
- $2\sqrt2$
- $4$
- $8$
Answer: (B) $2\sqrt2$
Standard deviation is the square root of variance, so $\sigma=\sqrt8=2\sqrt2$.
- 48
- 72
- 96
- 120
Answer: (B) 72
Total arrangements are $5!=120$. If the two specified books stay together, there are $2\times4!=48$ arrangements. Hence not together: $120-48=72$.
- $\frac{\sqrt{17}-3}{4}$
- $1/2$
- $1$
- $2$
Answer: (A) $\frac{\sqrt{17}-3}{4}$
Taking tangent gives $3x/(1-2x^2)=1$, so $2x^2+3x-1=0$. The positive root is $(\sqrt{17}-3)/4$.