JEE Main Full Mock Test: Practice Test 10
Question paper with answer key. Selection Pro
- $10\,\text{m/s}$
- $15\,\text{m/s}$
- $20\,\text{m/s}$
- $25\,\text{m/s}$
Answer: (C) $20\,\text{m/s}$
Using $v=u+at$, with $u=0$, $a=4\,\text{m/s}^2$ and $t=5\,\text{s}$, we get $v=20\,\text{m/s}$.
- $5\,\text{m}$
- $10\,\text{m}$
- $15\,\text{m}$
- $20\,\text{m}$
Answer: (B) $10\,\text{m}$
$H=\frac{u^2\sin^2\theta}{2g}=\frac{400(1/2)}{20}=10\,\text{m}$.
- $2\,\text{m/s}^2$
- $5\,\text{m/s}^2$
- $10\,\text{m/s}^2$
- $20\,\text{m/s}^2$
Answer: (B) $5\,\text{m/s}^2$
From Newton's second law, $F=ma$. Hence $a=F/m=10/2=5\,\text{m/s}^2$.
- $2\,\text{m/s}^2$
- $4\,\text{m/s}^2$
- $5\,\text{m/s}^2$
- $10\,\text{m/s}^2$
Answer: (B) $4\,\text{m/s}^2$
Since the surface is frictionless, net force is $20\,\text{N}$. Thus $a=F/m=20/5=4\,\text{m/s}^2$.
- $3\,\text{J}$
- $6\,\text{J}$
- $9\,\text{J}$
- $18\,\text{J}$
Answer: (C) $9\,\text{J}$
$K=\frac12mv^2=\frac12(2)(3^2)=9\,\text{J}$.
- $2\,\text{J}$
- $15\,\text{J}$
- $50\,\text{J}$
- $100\,\text{J}$
Answer: (C) $50\,\text{J}$
$W=Fs\cos0^\circ=10\times5=50\,\text{J}$.
- $1\,\text{J}$
- $2\,\text{J}$
- $4\,\text{J}$
- $8\,\text{J}$
Answer: (B) $2\,\text{J}$
$U=\frac12kx^2=\frac12(100)(0.2)^2=2\,\text{J}$.
- $8\,\text{N}$
- $16\,\text{N}$
- $32\,\text{N}$
- $64\,\text{N}$
Answer: (C) $32\,\text{N}$
$F_c=mv^2/r=2(16)/1=32\,\text{N}$.
- $5.6\,\text{km/s}$
- $7.9\,\text{km/s}$
- $11.2\,\text{km/s}$
- $15.0\,\text{km/s}$
Answer: (C) $11.2\,\text{km/s}$
The escape velocity from Earth is $v_e=\sqrt{2GM/R}$, approximately $11.2\,\text{km/s}$.
- $F/2$
- $F/4$
- $2F$
- $4F$
Answer: (B) $F/4$
Since $F\propto1/r^2$, doubling $r$ makes the force $F/4$.
- $1\,\text{s}$
- $2\,\text{s}$
- $\pi\,\text{s}$
- $4\,\text{s}$
Answer: (B) $2\,\text{s}$
$T=2\pi\sqrt{L/g}=2\pi\sqrt{1/\pi^2}=2\,\text{s}$.

- $100\,\text{m/s}$
- $200\,\text{m/s}$
- $300\,\text{m/s}$
- $500\,\text{m/s}$
Answer: (C) $300\,\text{m/s}$
Wave speed is $v=f\lambda=500\times0.6=300\,\text{m/s}$.
- $r$
- $1/r$
- $1/r^2$
- $r^2$
Answer: (C) $1/r^2$
From Coulomb's law, $E=\frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}$, so $E\propto1/r^2$.
- $5\,\mu\text{C}$
- $25\,\mu\text{C}$
- $50\,\mu\text{C}$
- $100\,\mu\text{C}$
Answer: (C) $50\,\mu\text{C}$
$Q=CV=5\,\mu\text{F}\times10\,\text{V}=50\,\mu\text{C}$.
- $0.5\,\text{A}$
- $1\,\text{A}$
- $2\,\text{A}$
- $4\,\text{A}$
Answer: (C) $2\,\text{A}$
Using Ohm's law, $I=V/R=20/10=2\,\text{A}$.
- $1\,\Omega$
- $2\,\Omega$
- $3\,\Omega$
- $9\,\Omega$
Answer: (B) $2\,\Omega$
$\frac1R=\frac16+\frac13=\frac12$, hence $R=2\,\Omega$.
- Straight line
- Circular
- Parabolic
- Elliptical
Answer: (B) Circular
The magnetic force is always perpendicular to velocity and acts as centripetal force, producing circular motion.
- $\frac{\mu_0I}{2R}$
- $\frac{\mu_0NI}{2R}$
- $\frac{\mu_0NI}{R}$
- $\frac{\mu_0I}{NR}$
Answer: (B) $\frac{\mu_0NI}{2R}$
For an $N$-turn circular coil, $B=\frac{\mu_0NI}{2R}$ at the centre.
- $0.5\,\text{V}$
- $5\,\text{V}$
- $50\,\text{V}$
- $100\,\text{V}$
Answer: (B) $5\,\text{V}$
Average emf $=N\Delta\Phi/\Delta t=N A\Delta B/\Delta t=100(0.01)(0.5)/0.1=5\,\text{V}$.
- Intensity is decreased
- Frequency of incident light is increased
- Work function is increased only
- Area of metal is increased
Answer: (B) Frequency of incident light is increased
$K_{\max}=h\nu-\phi=eV_s$. Therefore, increasing frequency increases the stopping potential.
- $0.5$
- $1$
- $2$
- $18$
Answer: (B) $1$
Molar mass of water is $18\,\text{g/mol}$. Thus moles $=18/18=1$.
- Neutrons
- Protons
- Nucleons
- Neutrons + electrons
Answer: (B) Protons
Atomic number $Z$ is defined as the number of protons in the nucleus.
- $3s$
- $3p$
- $3d$
- $4s$
Answer: (A) $3s$
For the same principal shell, penetration follows $s>p>d>f$. Hence $3s$ has greater penetration than $3p$ and $3d$.
- Oxygen
- Nitrogen
- Fluorine
- Chlorine
Answer: (C) Fluorine
Fluorine has the highest electronegativity on the Pauling scale.
- $NH_3$
- $H_2O$
- $CO_2$
- $CH_4$
Answer: (C) $CO_2$
$CO_2$ has two electron domains around carbon and is linear with bond angle $180^\circ$.
- $1$
- $1.5$
- $2$
- $3$
Answer: (C) $2$
The bond order of oxygen molecule is $2$, corresponding to a double bond.
- Positive
- Negative
- Zero
- Always infinite
Answer: (B) Negative
An exothermic reaction releases heat, so its enthalpy change is negative: $\Delta H<0$.
- Reactants
- Products
- Equal reactants and products
- Only catalyst
Answer: (B) Products
A large $K_c$ indicates that products are strongly favoured at equilibrium.
- $0$
- $5$
- $7$
- $14$
Answer: (C) $7$
At $25^\circ\text{C}$, neutral water has $[H^+]=10^{-7}\,\text{M}$, so pH is $7$.
- $+3$
- $+4$
- $+6$
- $+7$
Answer: (C) $+6$
$2(+1)+2x+7(-2)=0$, giving $2x=12$ and $x=+6$.
- $0\,\text{V}$
- $1\,\text{V}$
- $-1\,\text{V}$
- $0.76\,\text{V}$
Answer: (A) $0\,\text{V}$
The standard hydrogen electrode is assigned a standard reduction potential of exactly $0\,\text{V}$.
- Cathode
- Anode
- Salt bridge
- Electrolyte only
Answer: (B) Anode
Oxidation always occurs at the anode, while reduction occurs at the cathode.
- $\text{mol L}^{-1}\text{s}^{-1}$
- $\text{s}^{-1}$
- $\text{L mol}^{-1}\text{s}^{-1}$
- $\text{mol}^{-1}\text{s}^{-1}$
Answer: (B) $\text{s}^{-1}$
For a first-order reaction, rate $=k[A]$, so the unit of $k$ is $\text{s}^{-1}$.
- Rate constant
- Initial concentration
- Temperature
- Activation energy
Answer: (B) Initial concentration
For a first-order reaction, $t_{1/2}=\frac{0.693}{k}$, so it is independent of initial concentration.
- Viscosity
- Osmotic pressure
- Surface tension
- Refractive index
Answer: (B) Osmotic pressure
Osmotic pressure depends on the number of solute particles and is a colligative property.
- $-2$
- $-1$
- $0$
- $+1$
Answer: (B) $-1$
In peroxides, oxygen has oxidation state $-1$.
- $Na_2O$
- $CO_2$
- $Al_2O_3$
- $SO_2$
Answer: (C) $Al_2O_3$
$Al_2O_3$ reacts with both acids and bases, so it is amphoteric.
- $F_2$
- $Cl_2$
- $Br_2$
- $I_2$
Answer: (A) $F_2$
Fluorine has the highest standard reduction potential among halogens and is therefore the strongest oxidising agent.
- Methanol
- Ethanol
- Propanol
- Ethanal
Answer: (B) Ethanol
$CH_3CH_2OH$ contains two carbon atoms and an alcohol group, so its IUPAC name is ethanol.
- 1-bromopropane
- 2-bromopropane
- 1,2-dibromopropane
- Propanol
Answer: (B) 2-bromopropane
In the absence of peroxide, HBr follows Markovnikov addition and gives 2-bromopropane as the major product.
- $PCC$
- $NaBH_4$
- $LiAlH_4$
- $H_2/Ni$
Answer: (A) $PCC$
PCC (pyridinium chlorochromate) can oxidise a primary alcohol to an aldehyde without generally proceeding to the carboxylic acid.
- Ethane
- Propanone
- Ethanal
- Benzene
Answer: (C) Ethanal
Aldehydes such as ethanal reduce Tollens' reagent and produce the characteristic silver mirror.
- $-OH$
- $-CHO$
- $-COOH$
- $-CO-$
Answer: (C) $-COOH$
Carboxylic acids contain the carboxyl functional group $-COOH$.
- $3$
- $4$
- $6$
- $8$
Answer: (C) $6$
Six $NH_3$ ligands are directly coordinated to cobalt, so the coordination number is $6$.
- $Cl^-$
- $F^-$
- $CN^-$
- $Br^-$
Answer: (C) $CN^-$
$CN^-$ is a strong-field ligand and produces large crystal-field splitting.
- $1$
- $\sqrt{2}$
- $2$
- $3$
Answer: (B) $\sqrt{2}$
$|z|=\sqrt{1^2+1^2}=\sqrt2$.
- $6$
- $12$
- $15$
- $20$
Answer: (C) $15$
The coefficient is ${}^{6}C_2=15$.
- $8$
- $12$
- $16$
- $32$
Answer: (C) $16$
Equal roots require $b^2-4ac=0$. Thus $64-4k=0$, giving $k=16$.
- $190$
- $200$
- $210$
- $220$
Answer: (C) $210$
$S=\frac{n(n+1)}2=\frac{20\times21}{2}=210$.
- $8$
- $10$
- $12$
- $14$
Answer: (B) $10$
Determinant $=3(4)-2(1)=12-2=10$.
- $0^\circ$
- $45^\circ$
- $90^\circ$
- $180^\circ$
Answer: (A) $0^\circ$
Since $\vec a\cdot\vec b=|a||b|\cos\theta$, we have $\cos\theta=1$, giving $\theta=0^\circ$.
- $1$
- $2$
- $3$
- $4$
Answer: (B) $2$
$m=\frac{11-3}{6-2}=\frac84=2$.
- $y=3x+1$
- $y=3x-1$
- $y=x+3$
- $y=2x+3$
Answer: (B) $y=3x-1$
Using point-slope form, $y-2=3(x-1)$, hence $y=3x-1$.
- $\cos x$
- $-\cos x$
- $\sin x$
- $-\sin x$
Answer: (A) $\cos x$
$\frac{d}{dx}(\sin x)=\cos x$.
- $2$
- $4$
- $6$
- $8$
Answer: (B) $4$
$4x-x^2=4-(x-2)^2$. Therefore the maximum value is $4$.
- $0$
- $1$
- $2$
- $\pi$
Answer: (C) $2$
$\int_0^\pi\sin x\,dx=[-\cos x]_0^\pi=2$.
- $0$
- $1$
- $\infty$
- $-1$
Answer: (B) $1$
The standard trigonometric limit is $\lim_{x\to0}\frac{\sin x}{x}=1$.
- $3$
- $4$
- $5$
- $7$
Answer: (C) $5$
$d=\sqrt{3^2+4^2}=5$.
- $x^2+y^2=5$
- $x^2+y^2=10$
- $x^2+y^2=25$
- $x^2+y^2=50$
Answer: (C) $x^2+y^2=25$
For centre $(0,0)$ and radius $r$, the equation is $x^2+y^2=r^2$. Hence $x^2+y^2=25$.
- $0
- $e=1$
- $e>1$
- $e=2$
Answer: (A) $0
For an ellipse, eccentricity satisfies $0\le e<1$. For a non-circular ellipse, $0
- $\pi/6$
- $\pi/4$
- $\pi/3$
- $\pi/2$
Answer: (C) $\pi/3$
In the first quadrant, $\tan\theta=\sqrt3$ gives $\theta=\pi/3$.
- $0$
- $\pi/6$
- $\pi/4$
- $\pi/2$
Answer: (D) $\pi/2$
The principal range of $\sin^{-1}x$ is $[-\pi/2,\pi/2]$, hence $\sin^{-1}(1)=\pi/2$.
- $1/2$
- $1/3$
- $1/6$
- $1/12$
Answer: (C) $1/6$
There is one favourable outcome out of six equally likely outcomes, so probability is $1/6$.
- $4$
- $5$
- $6$
- $8$
Answer: (C) $6$
Mean $=\frac{2+4+6+8+10}{5}=\frac{30}{5}=6$.
- $0$
- $1$
- $5$
- $25$
Answer: (A) $0$
All observations are equal to their mean, so every deviation is zero. Hence variance is $0$.
- $6$
- $9$
- $12$
- $24$
Answer: (C) $12$
For a $2\times2$ matrix, $|kA|=k^2|A|$. Thus $|2A|=4(3)=12$.
- $y=x^2+C$
- $y=2x^2+C$
- $y=x+C$
- $y=2x+C$
Answer: (A) $y=x^2+C$
Integrating, $y=\int2x\,dx=x^2+C$.
- $1/4$
- $1/3$
- $1/2$
- $1$
Answer: (B) $1/3$
$\int_0^1x^2dx=\left[\frac{x^3}{3}\right]_0^1=\frac13$.
- $5$
- $7$
- $10$
- $14$
Answer: (C) $10$
For $ax^2+bx+c=0$, product of roots is $\alpha\beta=c/a$. Hence $\alpha\beta=10$.
- $14$
- $21$
- $28$
- $35$
Answer: (B) $21$
The number of selections is ${}^{7}C_2=\frac{7\times6}{2}=21$.
- $1$
- $2$
- $3$
- $10$
Answer: (C) $3$
Since $10^3=1000$, $\log_{10}1000=3$.
- $3$
- $4$
- $5$
- $6$
Answer: (C) $5$
$f'(x)=2x+3$. Therefore $f'(1)=2+3=5$.
- $e^x+C$
- $xe^x+C$
- $e^{x+1}+C$
- $x^2e^x+C$
Answer: (A) $e^x+C$
Since the derivative of $e^x$ is itself, $\int e^x dx=e^x+C$.
- $0$
- $1/2$
- $1$
- $2$
Answer: (B) $1/2$
Using the standard limit, $\lim_{x\to0}\frac{1-\cos x}{x^2}=\frac12$.
- $|a||b|$
- $1$
- $0$
- $-1$
Answer: (C) $0$
$|\vec a\times\vec b|=|a||b|\sin\theta$. For parallel vectors, $\theta=0^\circ$ or $180^\circ$, so the cross product is zero.