JEE Main Full Mock Test: Practice Test 08
Question paper with answer key. Selection Pro
- $x/2$
- $\sqrt{3}x/2$
- $x/\sqrt{2}$
- $x/4$
Answer: (B) $\sqrt{3}x/2$
For SHM, $v^2=\omega^2(A^2-y^2)$. With $v= v_{max}/2=\omega A/2$, we get $y^2=3A^2/4$, hence $y=\sqrt3A/2$.
- $2r$
- $4r$
- $r/2$
- $r$
Answer: (A) $2r$
Since $r=\sqrt{2mK}/qB$, radius is proportional to $\sqrt K$. Quadrupling kinetic energy doubles the radius.
- $U/2$
- $2U$
- $U$
- $4U$
Answer: (B) $2U$
After disconnection charge remains constant. Since $C\propto1/d$, doubling separation halves capacitance. As $U=Q^2/(2C)$, energy doubles.
- $30^\circ$
- $45^\circ$
- $\tan^{-1}(4/3)$
- $60^\circ$
Answer: (C) $\tan^{-1}(4/3)$
For a projectile, $H=\frac{u^2\sin^2\theta}{2g}$ and $R=\frac{u^2\sin2\theta}{g}$. Hence $\frac{H}{R}=\frac{\tan\theta}{4}$. Given $H=R/3$, $\tan\theta/4=1/3$, so $\tan\theta=4/3$.
- $1:2$
- $2:1$
- $1:4$
- $4:1$
Answer: (B) $2:1$
In parallel, voltage is same and $P=V^2/R$. Therefore $P_R:P_{2R}=1/R:1/(2R)=2:1$.
- $\sqrt{3g/L}$
- $\sqrt{g/L}$
- $\sqrt{6g/L}$
- $\sqrt{2g/L}$
Answer: (A) $\sqrt{3g/L}$
The centre of mass falls through $L/2$, so loss of gravitational potential energy is $mgL/2$. At the vertical position, $\frac12 I\omega^2=mgL/2$, with $I=\frac13mL^2$. Thus $\frac16mL^2\omega^2=\frac12mgL$, giving $\omega=\sqrt{3g/L}$.
- $1:3$
- $3:1$
- $1:1$
- $4:1$
Answer: (B) $3:1$
Potential energy fraction is $x^2/A^2=1/4$, so kinetic energy is $3/4$. Hence K:U=3:1.
- Negative
- Zero
- Positive
- Always equal to heat rejected
Answer: (C) Positive
For a clockwise cycle, the area enclosed in the P-V diagram is positive work done by the gas.
- $1:2$
- $2:1$
- $1:4$
- $4:1$
Answer: (B) $2:1$
Orbital speed $v=\sqrt{GM/r}$. Thus $v_R/v_{4R}=2$.
- Half
- Double
- Four times
- Unchanged
Answer: (B) Double
Fringe width $\beta=\lambda D/d$, so doubling wavelength doubles fringe width.
- $\sqrt{m_p/m_e}$
- $\sqrt{m_e/m_p}$
- $m_p/m_e$
- $1$
Answer: (A) $\sqrt{m_p/m_e}$
For the same accelerating potential, $\lambda=h/\sqrt{2mqV}\propto1/\sqrt m$. Hence $\lambda_e/\lambda_p=\sqrt{m_p/m_e}$.
- $R/n$
- $R/n^2$
- $nR$
- $R$
Answer: (B) $R/n^2$
Each part has resistance R/n. n identical pieces in parallel give $(R/n)/n=R/n^2$.
- $E/B$
- $B/E$
- $EB$
- $E+B$
Answer: (A) $E/B$
For crossed fields and perpendicular velocity, $qE=qvB$, hence $v=E/B$.
- 60 cm
- -60 cm
- 12 cm
- -12 cm
Answer: (A) 60 cm
$1/F=1/20-1/30=1/60$, hence F=60 cm.
- T
- $\sqrt T$
- $1/T$
- $T^2$
Answer: (B) $\sqrt T$
From kinetic theory, $v_{rms}=\sqrt{3RT/M}$, so it is proportional to $\sqrt T$.
- $MB$
- $M/B$
- $B/M$
- $0$
Answer: (A) $MB$
Torque is $\tau=MB\sin\theta$, maximum at $90^\circ$.
- $1/2$
- $1/4$
- $1/8$
- $1/16$
Answer: (C) $1/8$
30 days corresponds to three half-lives, so remaining fraction is $(1/2)^3=1/8$.
- 2 eV
- 3 eV
- 5 eV
- 7 eV
Answer: (B) 3 eV
Einstein's equation gives $K_{max}=E_\gamma-\phi=5-2=3$ eV.
- $2\omega$
- $\omega/2$
- $4\omega$
- $\omega/4$
Answer: (B) $\omega/2$
$\omega=1/\sqrt{LC}$. Increasing C fourfold reduces angular frequency by factor 2.
- v
- v/2
- 0
- -v
Answer: (C) 0
In a perfectly elastic head-on collision between identical masses, velocities are exchanged. The first body comes to rest.
- Zero everywhere
- Constant non-zero
- Inversely proportional to r
- Maximum at center
Answer: (A) Zero everywhere
By Gauss's law, a Gaussian surface inside the shell encloses zero charge, so electric field is zero.
- 44 V
- 110 V
- 440 V
- 22 V
Answer: (A) 44 V
$V_s/V_p=N_s/N_p=200/1000$, so $V_s=44$ V.
- $u/2$
- $u\sqrt3/2$
- $u/\sqrt2$
- $u\sqrt{3}/4$
Answer: (B) $u\sqrt3/2$
Using $v^2=u^2-2gh$ and $H=u^2/(2g)$, at $h=H/4$, $v^2=u^2(1-1/4)=3u^2/4$.
- U
- 2U
- 4U
- U/2
Answer: (C) 4U
Spring energy $U=\frac12kx^2$, so doubling extension makes energy four times.
- 2V
- V/2
- 4V
- V
Answer: (B) V/2
Potential of isolated sphere is $V=kQ/R$. Doubling radius halves potential.
- 2.32
- 3.32
- 4.32
- 1.32
Answer: (B) 3.32
For 90% completion, 10% remains. Number of half-lives $n=\log_2 10\approx3.32$.
- $O_2$
- $O_2^+$
- $O_2^-$
- $O_2^{2-}$
Answer: (B) $O_2^+$
Removing an electron from antibonding $\pi^*$ increases bond order. Thus $O_2^+$ has bond order 2.5.
- 2
- 3
- 11
- 1
Answer: (B) 3
Strong acid completely dissociates, so $[H^+]=10^{-3}$ M and pH=3.
- $[Fe(CN)_6]^{3-}$
- $[Fe(CN)_6]^{4-}$
- $[FeF_6]^{3-}$
- $[CoF_6]^{3-}$
Answer: (B) $[Fe(CN)_6]^{4-}$
$Fe^{2+}$ is d6 and CN− is strong field, giving low-spin $t_{2g}^6$, which is diamagnetic.
- Methyl chloride
- Ethyl chloride
- tert-Butyl chloride
- Vinyl chloride
Answer: (C) tert-Butyl chloride
SN1 depends on carbocation stability. The tert-butyl carbocation is highly stabilized by hyperconjugation and inductive effects.
- 2-methyl-1-butene
- 2-methyl-2-butene
- 1-butene
- 2-butene
Answer: (B) 2-methyl-2-butene
Acid-catalyzed dehydration follows Zaitsev's rule, giving the more substituted 2-methyl-2-butene as major product.
- B
- C
- N
- O
Answer: (C) N
Nitrogen has a stable half-filled 2p3 configuration, giving higher first ionization enthalpy than oxygen.
- 0
- 1
- Greater than 1 always
- Less than 1 always
Answer: (B) 1
An ideal gas obeys PV=nRT exactly, so $Z=PV/nRT=1$.
- $Zn+Cu^{2+}\to Zn^{2+}+Cu$
- $Cl_2+2OH^-\to Cl^-+ClO^-+H_2O$
- $NaCl\to Na^++Cl^-$
- $HCl+NaOH\to NaCl+H_2O$
Answer: (B) $Cl_2+2OH^-\to Cl^-+ClO^-+H_2O$
Chlorine is simultaneously reduced to Cl− and oxidized to ClO−, so it is disproportionation.
- 6 and 6
- 12 and 3
- 12 and 6
- 6 and 3
Answer: (B) 12 and 3
Benzene has 6 C-C sigma, 6 C-H sigma and 3 pi bonds: 12 sigma and 3 pi.
- Elevation in boiling point
- Depression in freezing point
- Osmotic pressure
- Relative lowering of vapour pressure
Answer: (C) Osmotic pressure
Proteins are high-molar-mass substances; osmotic pressure can be measured at room temperature using dilute solutions and is highly sensitive.
- Positive
- Zero only
- Negative
- Always equal to enthalpy change
Answer: (C) Negative
A spontaneous process at constant T and P has $\Delta G<0$.
- PCC
- $NaBH_4$
- $KMnO_4$
- $HNO_3$
Answer: (B) $NaBH_4$
Sodium borohydride reduces aldehydes to primary alcohols.
- sp
- sp2
- sp3
- dsp2
Answer: (A) sp
CO2 is linear with two electron domains around carbon, so carbon is sp hybridized.
- $Na_2O$
- $MgO$
- $Al_2O_3$
- $SO_3$
Answer: (C) $Al_2O_3$
Al2O3 reacts with both acids and bases, hence it is amphoteric.
- +3
- +4
- +6
- +7
Answer: (C) +6
$2(+1)+2x+7(-2)=0$ gives x=+6.
- LiF
- LiCl
- NaF
- KF
Answer: (A) LiF
Lattice enthalpy increases as ionic sizes decrease. Li+ and F− are the smallest pair here.
- 2-bromopropane
- 1-bromopropane
- Propane
- 1,2-dibromopropane
Answer: (B) 1-bromopropane
Peroxide effect causes anti-Markovnikov addition of HBr, giving 1-bromopropane.
- CO2
- BF3
- NH3
- CCl4
Answer: (C) NH3
CO2, BF3 and CCl4 are symmetrical and have zero net dipole moment. NH3 is polar.
- Increase
- Decrease
- Remain unchanged
- Become zero
Answer: (B) Decrease
For a weak electrolyte, degree of ionization decreases as concentration increases, approximately following Ostwald's dilution law.
- Polythene
- PVC
- Nylon-6,6
- Teflon
Answer: (C) Nylon-6,6
Nylon-6,6 is formed by condensation of hexamethylenediamine and adipic acid with elimination of water.
- Phenol
- Ethanol
- p-Nitrophenol
- p-Cresol
Answer: (C) p-Nitrophenol
The electron-withdrawing nitro group stabilizes the phenoxide ion, increasing acidity.
- Principal quantum number
- Azimuthal quantum number
- Magnetic quantum number
- Spin quantum number
Answer: (C) Magnetic quantum number
The magnetic quantum number m_l determines orbital orientation in space.
- Equal to pKa
- Equal to 7 always
- Equal to pKb
- Always zero
Answer: (A) Equal to pKa
Henderson-Hasselbalch equation gives pH=pKa+log([salt]/[acid]); equal concentrations make the logarithmic term zero.
- Copper
- Sodium
- Iron
- Zinc
Answer: (B) Sodium
Highly reactive sodium cannot be reduced from aqueous solution and is obtained by electrolysis of molten NaCl.
- 6
- 8
- 9
- 12
Answer: (A) 6
Let the roots be $a$ and $2a$. Their sum is $3a=6$, so $a=2$. Their product is $k=a(2a)=4$. Hence $k=4$.
- 1
- 1/2
- 1/4
- 2
Answer: (C) 1/4
$x(1-x)=1/4-(x-1/2)^2$, so maximum is 1/4.
- Real axis
- Imaginary axis
- Unit circle
- Line y=x
Answer: (B) Imaginary axis
Equal distances from 1 and -1 imply the perpendicular bisector of the segment joining them, i.e. the imaginary axis.
- 0
- 1
- e
- Infinity
Answer: (C) e
This is the standard exponential limit, equal to e.
- 1/3
- 1/6
- 1/2
- 2/3
Answer: (B) 1/6
Intersections are 0 and 1. Area is $\int_0^1(x-x^2)dx=1/2-1/3=1/6$.
- 1
- 5
- 7
- 12
Answer: (B) 5
For perpendicular vectors, $|a+b|=\sqrt{a^2+b^2}=5$.
- 5/7
- 15/7
- 3
- 15
Answer: (B) 15/7
Distance is $|10-(-5)|/\sqrt{4+9+36}=15/7$.
- Less than 1
- Equal to 1
- Greater than 1
- Zero
Answer: (C) Greater than 1
For a hyperbola, $e>1$.
- 0
- 1
- 2
- 3
Answer: (C) 2
The convex function e^x intersects the line x+2 twice: one root near -1.84 and another near 1.15.
- 6
- 18
- 54
- 9
Answer: (C) 54
For a $3\times3$ matrix, $|kA|=k^3|A|$. Therefore $|3A|=3^3\times2=54$.
- $-6$
- $-2$
- $2$
- $10$
Answer: (A) $-6$
$(1+x)^5(1-x)^3=(1+x)^2(1-x^2)^3$. Up to the $x^3$ term, $(1+2x+x^2)(1-3x^2+\cdots)$ gives the $x^3$ coefficient $2(-3)=-6$.
- $y=x^3+C$
- $y=3x+C$
- $y=x^2+C$
- $y=3x^3+C$
Answer: (A) $y=x^3+C$
Integrating $3x^2$ gives $y=x^3+C$.
- 1
- 5/6
- 6/5
- 1/5
Answer: (A) 1
$\tan(A+B)=(1/2+1/3)/(1-1/6)=1$.
- 1/2
- 2/3
- 67/100
- 5/6
Answer: (C) 67/100
Multiples of 2:50, of 3:33, of 6:16. Union=50+33-16=67.
- 2
- 4
- 6
- 8
Answer: (A) 2
Expanding the determinant gives $1(18-12)-1(9-3)+1(4-2)=6-6+2=2$. Hence the determinant is $2$.
- 3
- 4
- 5
- 6
Answer: (C) 5
Completing squares gives $(x-3)^2+(y+4)^2=25$, so radius is 5.
- $x^{x-1}$
- $x^x$
- $x^x(1+\ln x)$
- $x^{x}(\ln x-1)$
Answer: (C) $x^x(1+\ln x)$
Taking logs, ln y=x ln x. Hence y'/y=ln x+1, giving $y'=x^x(1+\ln x)$.
- 4 and 16
- 5 and 15
- 8 and 12
- 6 and 14
Answer: (A) 4 and 16
Sum=20 and product=64. The roots of t^2-20t+64=0 are 4 and 16.
- 1
- 2
- 3
- 4
Answer: (B) 2
x=0 is one solution and there is one positive nonzero intersection in (0,2), while no solution occurs beyond 2 because x/2>1.
- 1
- 2
- 3
- 7
Answer: (C) 3
Complete square: $(x+2)^2+3$, so minimum is 3.
- (2,-1,2)
- (2/3,-1/3,2/3)
- (1,-1,1)
- (1/2,-1,2)
Answer: (B) (2/3,-1/3,2/3)
Magnitude of direction-ratio vector is 3, so divide each component by 3.
- $\pi/2$
- $\pi/4$
- 1
- 2
Answer: (B) $\pi/4$
Antiderivative is tan^{-1}x, giving $\tan^{-1}(1)-\tan^{-1}(0)=\pi/4$.
- p>0,q<0
- p<0,q>0
- p>0,q>0
- p<0,q<0
Answer: (B) p<0,q>0
Sum of roots=-p>0 implies p<0; product=q>0. Thus p<0 and q>0.
- 4
- 8
- 16
- 10
Answer: (B) 8
Mean is 6. Squared deviations are 16,4,0,4,16; their average is 40/5=8.
- $ty=x+at^2$
- $ty=x+at^2$ with a negative sign
- $y=tx+a$
- $ty=x-at^2$
Answer: (A) $ty=x+at^2$
For the parabola $y^2=4ax$, the tangent at $(at^2,2at)$ is $ty=x+at^2$. Hence the correct option is $ty=x+at^2$.