JEE Main Full Mock Test: Practice Test 11
Question paper with answer key. Selection Pro
- $10\,\text{m}$
- $18\,\text{m}$
- $20\,\text{m}$
- $22\,\text{m}$
Answer: (B) $18\,\text{m}$
Distance covered in the $n$th second is $s_n=u+\frac{a}{2}(2n-1)$. For $u=0$, $a=4$ and $n=5$, $s_5=2(9)=18\,\text{m}$.
- $10\,\text{m}$
- $15\,\text{m}$
- $20\,\text{m}$
- $30\,\text{m}$
Answer: (B) $15\,\text{m}$
$H=\frac{u^2\sin^2\theta}{2g}=\frac{400\times3/4}{20}=15\,\text{m}$.
- $6\,\text{J}$
- $12\,\text{J}$
- $18\,\text{J}$
- $24\,\text{J}$
Answer: (B) $12\,\text{J}$
$W=\int_0^2 6x\,dx=3x^2\big|_0^2=12\,\text{J}$. Therefore the correct option is $12\,\text{J}$.
- $8\,\text{m/s}$
- $10\,\text{m/s}$
- $13\,\text{m/s}$
- $15\,\text{m/s}$
Answer: (C) $13\,\text{m/s}$
Acceleration is $a=F/m=4/2=2\,\text{m/s}^2$. Hence $v=u+at=3+2(5)=13\,\text{m/s}$.
- $2\,\text{m/s}^2$
- $4\,\text{m/s}^2$
- $8\,\text{m/s}^2$
- $16\,\text{m/s}^2$
Answer: (C) $8\,\text{m/s}^2$
Centripetal acceleration is $a_c=v^2/r=16/2=8\,\text{m/s}^2$.
- $\frac{1}{12}ML^2$
- $\frac{1}{6}ML^2$
- $\frac{1}{3}ML^2$
- $ML^2$
Answer: (C) $\frac{1}{3}ML^2$
For a uniform rod about an axis through one end perpendicular to its length, $I=\frac13ML^2$.
- Solid sphere
- Hollow sphere
- Both equal
- Cannot be determined
Answer: (B) Hollow sphere
For rolling, $K_{rot}=\frac12I\omega^2$. A hollow sphere has $I=\frac23MR^2$, whereas a solid sphere has $I=\frac25MR^2$, so the hollow sphere has greater rotational kinetic energy.
- $v_e$
- $v_e/\sqrt2$
- $\sqrt2v_e$
- $2v_e$
Answer: (B) $v_e/\sqrt2$
$v_e=\sqrt{2GM/R}$. On replacing $R$ by $2R$, $v'_e=v_e/\sqrt2$.
- $r/3$
- $r/2$
- $2r/3$
- $3r/4$
Answer: (A) $r/3$
Let the point be at distance $x$ from $q$. Equating field magnitudes, $kq/x^2=4kq/(r-x)^2$. Thus $r-x=2x$, giving $x=r/3$.
- $U/2$
- $U$
- $2U$
- $4U$
Answer: (C) $2U$
For constant voltage, $U=\frac12CV^2$. Doubling $C$ doubles the stored energy.
- $R/4$
- $R/2$
- $2R$
- $4R$
Answer: (D) $4R$
Since volume is constant, doubling length makes area half. As $R=\rho L/A$, the resistance becomes $2R/(1/2)=4R$.
- Circular
- Helical
- Straight line
- Parabolic
Answer: (C) Straight line
Magnetic force is $qvB\sin\theta$. For motion parallel to the field, $\theta=0$, so the force is zero and the particle continues in a straight line.
- 2 times
- 4 times
- 6 times
- 8 times
Answer: (B) 4 times
Magnetic moment is $M=NIA$. Doubling both $N$ and $I$ makes it four times.
- $22\,\text{V}$
- $44\,\text{V}$
- $110\,\text{V}$
- $440\,\text{V}$
Answer: (B) $44\,\text{V}$
For an ideal transformer, $V_s/V_p=N_s/N_p=100/500=1/5$. Hence $V_s=44\,\text{V}$.
- Parallel to each other
- Perpendicular to each other and to the direction of propagation
- Only parallel to propagation
- Randomly oriented
Answer: (B) Perpendicular to each other and to the direction of propagation
Electromagnetic waves are transverse. The electric field, magnetic field and direction of propagation are mutually perpendicular.
- $10\,\text{cm}$
- $15\,\text{cm}$
- $20\,\text{cm}$
- $30\,\text{cm}$
Answer: (B) $15\,\text{cm}$
Using the lens formula $1/f=1/v-1/u$, with $f=10\,\text{cm}$ and $u=-30\,\text{cm}$, we get $1/v=1/10-1/30=1/15$. Thus $v=15\,\text{cm}$.
- $\beta/2$
- $\beta$
- $2\beta$
- $4\beta$
Answer: (A) $\beta/2$
Fringe width is $\beta=\lambda D/d$. Doubling $d$ makes the fringe width half.
- $3\times10^{14}h$
- $5\times10^{14}h$
- $8\times10^{14}h$
- $13\times10^{14}h$
Answer: (A) $3\times10^{14}h$
Einstein's equation gives $K_{max}=h(\nu-\nu_0)=h(8-5)\times10^{14}=3\times10^{14}h$.
- $\lambda$
- $1/\lambda$
- $\lambda^2$
- $1/\lambda^2$
Answer: (B) $1/\lambda$
The mean life of a radioactive nucleus is $\tau=1/\lambda$.
- $-3.4\,\text{eV}$
- $-6.8\,\text{eV}$
- $-13.6\,\text{eV}$
- $+13.6\,\text{eV}$
Answer: (C) $-13.6\,\text{eV}$
For hydrogen, energy of the $n$th orbit is $E_n=-13.6/n^2\,\text{eV}$. For $n=1$, $E=-13.6\,\text{eV}$.
- Work done
- Change in internal energy
- Heat exchanged
- Pressure
Answer: (C) Heat exchanged
In an adiabatic process, there is no heat exchange, so $Q=0$.
- Half
- Unchanged
- Twice
- Four times
Answer: (C) Twice
RMS speed varies as $v_{rms}\propto\sqrt T$. Thus when $T$ becomes $4T$, speed becomes $2v_{rms}$.
- $4/3$
- $5/3$
- $7/5$
- $3/2$
Answer: (B) $5/3$
For a monoatomic ideal gas, $C_V=3R/2$ and $C_P=5R/2$, hence $\gamma=5/3$.
- $R/2$
- $R/4$
- $R/8$
- $2R$
Answer: (A) $R/2$
Volume is conserved: $\frac43\pi R^3=8\times\frac43\pi r^3$. Thus $r^3=R^3/8$ and $r=R/2$.
- $1\,\text{s}$
- $2\,\text{s}$
- $4\,\text{s}$
- $8\,\text{s}$
Answer: (C) $4\,\text{s}$
Since $T\propto1/\sqrt g$, reducing $g$ to $g/4$ makes the period twice as large. Hence $T'=4\,\text{s}$.
- $3.011\times10^{23}$
- $6.022\times10^{23}$
- $1.204\times10^{24}$
- $9.033\times10^{23}$
Answer: (B) $6.022\times10^{23}$
$18\,\text{g}$ of water is one mole. Therefore it contains Avogadro's number, $6.022\times10^{23}$ molecules.
- Mass only
- Momentum
- Kinetic energy only
- Charge
Answer: (B) Momentum
According to de Broglie's relation, $\lambda=h/p$, so wavelength is inversely proportional to momentum.
- $\Delta G>0$
- $\Delta G=0$ always
- $\Delta G<0$
- $\Delta G=\Delta H$ always
Answer: (C) $\Delta G<0$
A spontaneous process at constant $T$ and $P$ has $\Delta G<0$.
- Mostly reactants
- Mostly products
- Only reactants
- No products
Answer: (B) Mostly products
A large equilibrium constant indicates that products are favoured at equilibrium.
- $2$
- $3$
- $4$
- $10$
Answer: (C) $4$
For a strong acid, $[H^+]=10^{-4}\,\text{M}$, so pH $=4$.
- $+3$
- $+4$
- $+6$
- $+7$
Answer: (C) $+6$
$2(+1)+2x+7(-2)=0$, giving $2x=12$ and $x=+6$.
- Li
- Be
- B
- C
Answer: (D) C
First ionisation enthalpy generally increases across the period. Among Li, Be, B and C, carbon has the highest value.
- $NH_3$
- $H_2O$
- $CO_2$
- $CH_4$
Answer: (C) $CO_2$
$CO_2$ has two electron domains around carbon and no lone pair on the central atom, giving a linear geometry with bond angle $180^\circ$.
- NaCl
- KCl
- MgO
- CaO
Answer: (C) MgO
MgO contains $Mg^{2+}$ and $O^{2-}$ ions, giving much stronger electrostatic attraction than the singly charged ions in NaCl and KCl.
- $\text{mol L}^{-1}\text{s}^{-1}$
- $\text{s}^{-1}$
- $\text{L mol}^{-1}\text{s}^{-1}$
- $\text{mol}^{-1}\text{s}^{-1}$
Answer: (B) $\text{s}^{-1}$
For a first-order reaction, rate $=k[A]$, so $k$ has unit time$^{-1}$, i.e. $\text{s}^{-1}$.
- $0.0693\,\text{min}^{-1}$
- $0.693\,\text{min}^{-1}$
- $6.93\,\text{min}^{-1}$
- $0.00693\,\text{min}^{-1}$
Answer: (A) $0.0693\,\text{min}^{-1}$
For a first-order reaction, $k=0.693/t_{1/2}=0.693/10=0.0693\,\text{min}^{-1}$.
- Negative
- Zero
- Positive
- Always one
Answer: (C) Positive
$\Delta G^\circ=-nFE^\circ_{cell}$. For spontaneity, $\Delta G^\circ<0$, requiring $E^\circ_{cell}>0$.
- Anode
- Cathode
- Salt bridge
- Both electrodes
Answer: (B) Cathode
At the cathode, reduction occurs: $Na^+ + e^-\rightarrow Na$.
- 1-bromopropane
- 2-bromopropane
- 1,2-dibromopropane
- Propanol
Answer: (A) 1-bromopropane
In the presence of peroxide, HBr adds through a radical mechanism following anti-Markovnikov orientation, giving 1-bromopropane.
- Tollens' reagent
- NaCl
- HCl
- Conc. H2SO4
Answer: (A) Tollens' reagent
Aldehydes generally reduce Tollens' reagent to metallic silver, while ordinary ketones do not.
- Acetone
- Benzaldehyde
- Ethanal
- Benzophenone
Answer: (C) Ethanal
Aliphatic aldehydes such as ethanal reduce Fehling's solution to give a brick-red precipitate of $Cu_2O$.
- Ethene
- Propene
- 2-methylpropene
- 2-methylpropane
Answer: (C) 2-methylpropene
Acid-catalysed dehydration of tert-butanol gives 2-methylpropene as the alkene product.
- Tetrahedral
- Square planar
- Octahedral
- Linear
Answer: (B) Square planar
$Pt^{2+}$ is a $d^8$ ion and commonly forms square-planar complexes such as $[PtCl_4]^{2-}$.
- 4
- 5
- 6
- 8
Answer: (C) 6
Six cyanide ligands are directly coordinated to chromium, so the coordination number is $6$.
- Nylon-6,6
- Terylene
- PVC
- Bakelite
Answer: (C) PVC
PVC is formed by addition polymerisation of vinyl chloride molecules.
- Ethene
- Vinyl chloride
- Tetrafluoroethene
- Styrene
Answer: (C) Tetrafluoroethene
Teflon (PTFE) is obtained by polymerisation of tetrafluoroethene, $CF_2=CF_2$.
- Ethanol
- Phenol
- Water
- Acetic acid
Answer: (D) Acetic acid
Acetic acid has a resonance-stabilised carboxylate conjugate base and is substantially more acidic than phenol, water and ethanol.
- $6,3$
- $12,3$
- $12,6$
- $6,6$
Answer: (B) $12,3$
Benzene contains six C-C sigma bonds and six C-H sigma bonds, giving 12 sigma bonds. It has three pi bonds in its Kekule representation.
- $O_2$
- $O_2^+$
- $O_2^-$
- $O_2^{2-}$
Answer: (B) $O_2^+$
Removing one electron from antibonding $\pi^*$ orbitals increases the bond order. Thus $O_2^+$ has bond order $2.5$, greater than $O_2$ ($2$), $O_2^-$ ($1.5$) and $O_2^{2-}$ ($1$).
- $F_2$
- $Cl_2$
- $Li$
- $O_2$
Answer: (C) $Li$
Lithium readily loses its valence electron and has a very negative standard reduction potential, making it a strong reducing agent.
- $-4$
- $4$
- $-4i$
- $4i$
Answer: (A) $-4$
$(1-i)^2=-2i$, so $(1-i)^4=(-2i)^2=-4$.
- $84$
- $126$
- $144$
- $210$
Answer: (B) $126$
The coefficient is ${}^{9}C_4=\frac{9\times8\times7\times6}{24}=126$.
- $10$
- $12$
- $13$
- $14$
Answer: (B) $12$
Let roots be $\alpha,\beta$. Then $(\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta$. Thus $1=49-4k$, giving $k=12$. Therefore the correct option is $12$.
- $760$
- $820$
- $820$
- $840$
Answer: (B) $820$
Here $a=3$, $d=4$, $n=20$. $S_n=\frac n2[2a+(n-1)d]=10[6+76]=820$.
- $15$
- $25$
- $50$
- $75$
Answer: (D) $75$
For a $2\times2$ matrix, $|kA|=k^2|A|$. Hence $|5A|=25\times3=75$. Therefore the correct option is $75$.
- $-4$
- $0$
- $2$
- $4$
Answer: (B) $0$
$\vec a\cdot\vec b=(2)(1)+(1)(-2)=0$.
- $-3$
- $0$
- $3$
- $6$
Answer: (B) $0$
$dy/dx=3x^2-3$. At $x=1$, slope $=3-3=0$.
- $\frac{64}{27}$
- $\frac{128}{27}$
- $\frac{256}{27}$
- $8$
Answer: (C) $\frac{256}{27}$
Let $f(x)=4x^2-x^3$. Then $f'(x)=8x-3x^2=x(8-3x)=0 \implies x=8/3$. Maximum value $f(8/3)=(8/3)^2(4-8/3)=\frac{256}{27}$.
- $0$
- $1$
- $2$
- $\pi/2$
Answer: (B) $1$
$\int_0^{\pi/2}\sin x\,dx=[-\cos x]_0^{\pi/2}=1$.
- $\frac{1}{x^2+1}$
- $\frac{2x}{x^2+1}$
- $\frac{x}{x^2+1}$
- $2x\ln(x^2+1)$
Answer: (B) $\frac{2x}{x^2+1}$
Using the chain rule, $\frac{d}{dx}\ln(x^2+1)=\frac{2x}{x^2+1}$.
- $1$
- $2$
- $3$
- $0$
Answer: (C) $3$
$\frac{\sin3x}{x}=3\frac{\sin3x}{3x}\to3$.
- $2$
- $3$
- $4$
- $5$
Answer: (B) $3$
Distance $=|c_1-c_2|/\sqrt{a^2+b^2}=15/5=3$.
- $x^2+y^2-4x=0$
- $x^2+y^2+4x=0$
- $x^2+y^2-2x=0$
- $x^2+y^2+2x=0$
Answer: (A) $x^2+y^2-4x=0$
The centre is $(2,0)$ and radius is $2$. Thus $(x-2)^2+y^2=4$, giving $x^2+y^2-4x=0$.
- $2$
- $4$
- $8$
- $16$
Answer: (C) $8$
For $y^2=4ax$, $4a=8$, so $a=2$. Latus rectum length is $4a=8$.
- $4/3$
- $5/3$
- $3/5$
- $5/4$
Answer: (B) $5/3$
Here $a^2=9$ and $b^2=16$. Thus $c^2=a^2+b^2=25$, so $e=c/a=5/3$.
- $3/5$
- $4/5$
- $5/3$
- $1/5$
Answer: (B) $4/5$
For an acute angle, $\cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-9/25}=4/5$.
- $\pi/6$
- $-\pi/6$
- $5\pi/6$
- $-5\pi/6$
Answer: (B) $-\pi/6$
The principal range of $\sin^{-1}x$ is $[-\pi/2,\pi/2]$. Therefore $\sin^{-1}(-1/2)=-\pi/6$.
- $1/18$
- $1/9$
- $1/12$
- $5/36$
Answer: (B) $1/9$
Favourable outcomes are $(3,6),(4,5),(5,4),(6,3)$, four outcomes out of $36$. Probability $=4/36=1/9$.
- $4$
- $5$
- $6$
- $8$
Answer: (B) $5$
Mean $=5$. Squared deviations are $9,1,1,9$, whose sum is $20$. Population variance $=20/4=5$.
- $5$
- $-5$
- $1/5$
- $25$
Answer: (C) $1/5$
For a nonsingular matrix, $|A^{-1}|=1/|A|$. Hence $|A^{-1}|=1/5$.
- $y=2x+C$
- $y=Ce^{2x}$
- $y=Ce^x$
- $y=x^2+C$
Answer: (B) $y=Ce^{2x}$
Separating variables, $dy/y=2dx$. Integrating gives $\ln y=2x+C$, hence $y=Ce^{2x}$.
- $4/3$
- $8/3$
- $3$
- $4$
Answer: (B) $8/3$
$\int_0^2x^2dx=[x^3/3]_0^2=8/3$.
- $1$
- $2$
- $4$
- $5$
Answer: (C) $4$
$\frac1\alpha+\frac1\beta=\frac{\alpha+\beta}{\alpha\beta}=4/1=4$.
- $25$
- $60$
- $100$
- $120$
Answer: (D) $120$
The number of arrangements of $5$ distinct objects is $5!=120$.
- $14$
- $21$
- $28$
- $35$
Answer: (B) $21$
Number of selections is ${}^{7}C_2=7\times6/2=21$.