JEE Main Full Mock Test: Practice Test 09

1. $A/2$
2. $A/\sqrt{2}$
3. $A/\sqrt{3}$
4. $A/4$

Answer: (B) $A/\sqrt{2}$

Total energy is $E=\frac{1}{2}m\omega^2A^2$. Potential energy at displacement $x$ is $U=\frac{1}{2}m\omega^2x^2$. For $K=U$, each must be half of total energy. Thus $x^2=A^2/2$, giving $|x|=A/\sqrt{2}$.

1. $1/2$
2. $1$
3. $2$
4. $4$

Answer: (B) $1$

Maximum height is $H=\frac{u^2\sin^2\theta}{2g}$ and range is $R=\frac{u^2\sin2\theta}{g}$. Therefore $\frac{H}{R}=\frac{\tan\theta}{4}$. Given $H/R=1/4$, we get $\tan\theta=1$.

1. $F/k$
2. $2F/k$
3. $F/(2k)$
4. $\sqrt{F/k}$

Answer: (B) $2F/k$

At maximum extension, the block is momentarily at rest. Work done by the constant force becomes spring potential energy: $Fx=\frac{1}{2}kx^2$. Hence the non-zero solution is $x=2F/k$.

1. $0$
2. $v/2$
3. $v$
4. $2v$

Answer: (C) $v$

For a perfectly elastic head-on collision between identical masses, the velocities are exchanged. Therefore the second body moves with speed $v$.

1. $K$
2. $-K$
3. $2K$
4. $-2K$

Answer: (B) $-K$

For a circular orbit, $K=GMm/(2r)$ and potential energy is $U=-GMm/r=-2K$. Hence total energy $E=K+U=-K$.

1. $d/3$
2. $d/2$
3. $d/4$
4. $2d/3$

Answer: (A) $d/3$

Let the point be at distance $x$ from $q$. Equating field magnitudes, $kq/x^2=4kq/(d-x)^2$. Thus $d-x=2x$, giving $x=d/3$.

1. $\frac{1}{2}CV^2$
2. $\frac{1}{2}KCV^2$
3. $\frac{1}{2K}CV^2$
4. $K^2CV^2$

Answer: (C) $\frac{1}{2K}CV^2$

After disconnection, charge remains constant. The new capacitance is $KC$. Hence $U'=Q^2/(2KC)$. Since initially $Q=CV$, we obtain $U'=CV^2/(2K)$.

1. $R/2$
2. $R$
3. $2R$
4. $4R$

Answer: (D) $4R$

Volume remains constant, so when length becomes $2L$, area becomes $A/2$. Since $R=\rho L/A$, the new resistance is $\rho(2L)/(A/2)=4R$.

1. A straight line
2. A circle
3. A parabola
4. A helix

Answer: (D) A helix

The velocity component perpendicular to the magnetic field produces circular motion, while the parallel component remains uniform. Their combination produces a helical path.

1. $BL/v$
2. $BvL$
3. $Bv/L$
4. $BLv^2$

Answer: (B) $BvL$

For a rod moving perpendicular to the magnetic field, motional emf is $\varepsilon=BLv$.

1. $R$
2. $\sqrt{R^2+(X_L+X_C)^2}$
3. $X_L-X_C$
4. $0$

Answer: (A) $R$

At resonance, $X_L=X_C$, so the net reactance is zero. Therefore the impedance becomes $Z=R$.

1. Virtual, erect and magnified
2. Real, inverted and same size
3. Real, erect and magnified
4. Virtual and diminished

Answer: (B) Real, inverted and same size

For an object at $2f$ of a convex lens, the image forms at $2f$ on the opposite side. It is real, inverted and of the same size.

1. Half
2. Unchanged
3. Double
4. Four times

Answer: (C) Double

Fringe width is $\beta=\lambda D/d$. Therefore, doubling the wavelength doubles the fringe width.

1. $1\,eV$
2. $2\,eV$
3. $3\,eV$
4. $5\,eV$

Answer: (B) $2\,eV$

Photon energy is $E=hc/\lambda=1240/248=5\,eV$. Therefore $K_{max}=5-3=2\,eV$.

1. $\sqrt{m_e/m_p}$
2. $\sqrt{m_p/m_e}$
3. $m_p/m_e$
4. $1$

Answer: (B) $\sqrt{m_p/m_e}$

For a charged particle accelerated through the same potential difference, $\lambda=h/\sqrt{2mqV}$, so $\lambda\propto1/\sqrt m$. Hence $\lambda_e/\lambda_p=\sqrt{m_p/m_e}$.

1. $\lambda$
2. $1/\lambda$
3. $\lambda^2$
4. $\ln2/\lambda$

Answer: (B) $1/\lambda$

The mean life of a radioactive substance is $\tau=1/\lambda$.

1. $n$
2. $1/n$
3. $n^2$
4. $1/n^2$

Answer: (C) $n^2$

For a hydrogen atom in Bohr's model, the orbit radius varies as $r_n\propto n^2$.

1. $Q=0$
2. $W=0$
3. $\Delta U=0$
4. $T$ remains constant

Answer: (A) $Q=0$

An adiabatic process is defined as a process in which there is no heat exchange, so $Q=0$.

1. $T$
2. $1/T$
3. $\sqrt{T}$
4. $T^2$

Answer: (C) $\sqrt{T}$

The rms speed is $v_{rms}=\sqrt{3RT/M}$, hence $v_{rms}\propto\sqrt T$.

1. $h/4$
2. $h/2$
3. $2h$
4. $4h$

Answer: (B) $h/2$

Capillary rise is $h=\frac{2T\cos\theta}{\rho gr}$, so $h\propto1/r$. Doubling the radius halves the capillary rise.

1. $u^2/g$
2. $u^2/(2g)$
3. $2u^2/g$
4. $u/g$

Answer: (B) $u^2/(2g)$

Using $v^2=u^2-2gh$ and $v=0$ at maximum height gives $h=u^2/(2g)$.

1. $ML^2/12$
2. $ML^2/6$
3. $ML^2/3$
4. $ML^2$

Answer: (C) $ML^2/3$

For a uniform rod about an axis through one end perpendicular to its length, $I=ML^2/3$.

1. Hollow sphere
2. Solid sphere
3. Both simultaneously
4. Depends only on mass

Answer: (B) Solid sphere

Rolling acceleration is $a=g\sin\theta/[1+I/(MR^2)]$. For a solid sphere $I=2MR^2/5$, while for a thin hollow sphere $I=2MR^2/3$. The solid sphere therefore has greater acceleration.

1. $K_aC$
2. $\sqrt{K_aC}$
3. $K_a/C$
4. $C/K_a$

Answer: (B) $\sqrt{K_aC}$

For a weak acid, $K_a\approx x^2/C$, where $x=[H^+]$. Therefore $[H^+]\approx\sqrt{K_aC}$.

1. $5\,min$
2. $10\,min$
3. $20\,min$
4. $30\,min$

Answer: (B) $10\,min$

For a first-order reaction, $t_{1/2}=0.693/k=0.693/0.0693=10\,min$.

1. Products
2. Reactants
3. No change
4. Both equally

Answer: (B) Reactants

The reactant side has 3 moles of gas while the product side has 2. Increasing volume lowers pressure and favors the side with more gaseous moles, so equilibrium shifts toward reactants.

1. $\Delta G^\circ>0$
2. $\Delta G^\circ=0$
3. $\Delta G^\circ<0$
4. $K=0$

Answer: (C) $\Delta G^\circ<0$

The relation is $\Delta G^\circ=-nFE^\circ_{cell}$. Thus a positive standard cell potential corresponds to a negative standard Gibbs energy change.

1. $+3$
2. $+4$
3. $+6$
4. $+7$

Answer: (C) $+6$

Let the oxidation state of Cr be $x$. Then $2(+1)+2x+7(-2)=0$, giving $x=+6$.

1. $NaCl$
2. $KCl$
3. $MgO$
4. $CsCl$

Answer: (C) $MgO$

Lattice enthalpy magnitude increases with higher ionic charges and smaller ionic sizes. $MgO$ has $2+$ and $2-$ ions, giving stronger electrostatic attraction.

1. Tetrahedral
2. Square planar
3. Trigonal planar
4. Linear

Answer: (B) Square planar

XeF4 has six electron pairs around xenon: four bond pairs and two lone pairs. The lone pairs occupy opposite positions, giving a square-planar molecular shape.

1. $[FeF_6]^{3-}$
2. $[Fe(CN)_6]^{3-}$
3. $[Co(NH_3)_6]^{3+}$
4. $[NiCl_4]^{2-}$

Answer: (C) $[Co(NH_3)_6]^{3+}$

In $[Co(NH_3)_6]^{3+}$, cobalt is $Co^{3+}$ with a $d^6$ configuration. The octahedral complex is low spin, giving $t_{2g}^6$ with all electrons paired. Hence it is diamagnetic.

1. $4$
2. $5$
3. $6$
4. $8$

Answer: (C) $6$

Six cyanide ligands are directly coordinated to chromium, so the coordination number is $6$.

1. Methyl chloride
2. Ethyl chloride
3. Isopropyl chloride
4. tert-Butyl chloride

Answer: (D) tert-Butyl chloride

SN1 reactions involve carbocation formation. The tert-butyl carbocation is the most stable among the given carbocations, so tert-butyl chloride reacts fastest.

1. Markovnikov's rule
2. Saytzeff's rule
3. Anti-Markovnikov rule
4. Hund's rule

Answer: (B) Saytzeff's rule

Acid-catalysed dehydration of alcohols generally follows Saytzeff's rule, favoring the more substituted alkene.

1. Phenol has higher molecular mass
2. Phenoxide ion is resonance-stabilized
3. Ethanol is aromatic
4. Phenol contains more hydrogen

Answer: (B) Phenoxide ion is resonance-stabilized

The conjugate base of phenol, phenoxide ion, is stabilized by resonance. Ethoxide ion does not have comparable resonance stabilization.

1. Ethanol
2. Ethene
3. Ethanoic acid
4. Methanoic acid

Answer: (C) Ethanoic acid

Aldehydes are readily oxidized to carboxylic acids. Therefore ethanal gives ethanoic acid.

1. Tollens' reagent
2. NaCl
3. HCl
4. NaOH only

Answer: (A) Tollens' reagent

Aldehydes generally reduce Tollens' reagent to metallic silver, whereas ordinary ketones do not.

1. $CH_3COOH$
2. $ClCH_2COOH$
3. $Cl_2CHCOOH$
4. $Cl_3CCOOH$

Answer: (D) $Cl_3CCOOH$

Chlorine has a strong electron-withdrawing inductive effect. Increasing the number of chlorine atoms stabilizes the carboxylate ion and increases acidity.

1. Aniline has no lone pair
2. The lone pair on nitrogen in aniline is delocalized into the benzene ring
3. Methylamine is aromatic
4. Aniline has an acidic hydrogen

Answer: (B) The lone pair on nitrogen in aniline is delocalized into the benzene ring

In aniline, the nitrogen lone pair participates in resonance with the benzene ring, making it less available for protonation than the lone pair in methylamine.

1. Polythene
2. PVC
3. Nylon-6,6
4. Teflon

Answer: (C) Nylon-6,6

Nylon-6,6 is produced by condensation polymerisation of hexamethylenediamine and adipic acid with elimination of small molecules such as water.

1. $1$
2. $2$
3. $3$
4. $4$

Answer: (B) $2$

The square-planar complex can have cis and trans arrangements, giving two geometrical isomers.

1. Positive
2. Negative
3. Zero
4. Infinite

Answer: (C) Zero

For an ideal solution, interactions between unlike molecules are effectively equal to those between like molecules, so $\Delta H_{mix}=0$.

1. $p_i=p_i^\circ/x_i$
2. $p_i=x_i p_i^\circ$
3. $p_i=x_i/p_i^\circ$
4. $p_i=p_i^\circ+x_i$

Answer: (B) $p_i=x_i p_i^\circ$

Raoult's law states $p_i=x_i p_i^\circ$, where $x_i$ is the mole fraction and $p_i^\circ$ is the vapour pressure of pure component $i$.

1. At all temperatures
2. At no temperature
3. At sufficiently low temperature
4. Only at very high temperature

Answer: (C) At sufficiently low temperature

Spontaneity requires $\Delta G=\Delta H-T\Delta S<0$. When both $\Delta H$ and $\Delta S$ are negative, sufficiently low temperature favors spontaneity.

1. $H_3PO_4$
2. $HPO_4^{2-}$
3. $PO_4^{3-}$
4. $H_2PO_3^-$

Answer: (B) $HPO_4^{2-}$

Removal of one proton from $H_2PO_4^-$ produces $HPO_4^{2-}$, which is its conjugate base.

1. Na
2. Mg
3. Al
4. Si

Answer: (D) Si

Across the third period, first ionization enthalpy generally increases. Although Mg has higher ionization enthalpy than Al due to subshell configuration, Si has the highest among the given elements.

1. $Na_2O$
2. $MgO$
3. $Al_2O_3$
4. $SO_3$

Answer: (C) $Al_2O_3$

$Al_2O_3$ reacts with both acids and bases, so it is classified as an amphoteric oxide.

1. $1.5$
2. $2$
3. $2.5$
4. $3$

Answer: (C) $2.5$

The bond order of $O_2$ is 2. Removing one electron from an antibonding $\pi^*$ orbital increases the bond order by $1/2$. Thus the bond order of $O_2^+$ is $2.5$.

1. $x=0$
2. $y=0$
3. $x=2$
4. $y=2$

Answer: (A) $x=0$

The point $z$ is equidistant from $(2,0)$ and $(-2,0)$. Their perpendicular bisector is the imaginary axis, so $x=0$.

1. $-2$
2. $-5$
3. $6$
4. $8$

Answer: (A) $-2$

$(1+x)^6(1-x)^4=(1-x^2)^4(1+x)^2$. The $x^4$ coefficient comes from $6x^4$ in $(1-x^2)^4$ times $1$, and from $(-4x^2)(2x^2)$, giving $6-8=-2$.

1. $0$
2. $2$
3. $4$
4. $8$

Answer: (B) $2$

For equal roots, the discriminant must be zero: $(a+2)^2-8a=0$. Hence $a^2-4a+4=0$, giving $(a-2)^2=0$ and therefore $a=2$.

1. $6n-1$
2. $6n+1$
3. $3n+2$
4. $3n-1$

Answer: (A) $6n-1$

$a_n=S_n-S_{n-1}$. Therefore $a_n=(3n^2+2n)-[3(n-1)^2+2(n-1)]=6n-1$.

1. $3$
2. $1/3$
3. $-3$
4. $1/9$

Answer: (B) $1/3$

For a non-singular matrix, $|A^{-1}|=1/|A|$. Hence $|A^{-1}|=1/3$.

1. $30^\circ$
2. $45^\circ$
3. $60^\circ$
4. $90^\circ$

Answer: (C) $60^\circ$

$|a+b|^2=|a|^2+|b|^2+2a\cdot b=2+2\cos\theta$. Given $|a+b|^2=3$, we get $\cos\theta=1/2$, so $\theta=60^\circ$.

1. $ty=x+at^2$
2. $ty=x-at^2$
3. $y=tx+a$
4. $y=tx-at^2$

Answer: (A) $ty=x+at^2$

For the parabola $y^2=4ax$, the point corresponding to parameter $t$ is $(at^2,2at)$. Its tangent is $ty=x+at^2$.

1. $1$
2. $\sqrt2$
3. $2$
4. $1/\sqrt2$

Answer: (B) $\sqrt2$

$\sin x+\cos x=\sqrt2\sin(x+\pi/4)$. Since the maximum value of sine is $1$, the maximum value is $\sqrt2$.

1. $-1/6$
2. $1/6$
3. $-1/3$
4. $1/3$

Answer: (A) $-1/6$

Using the expansion $\sin x=x-x^3/6+O(x^5)$, we get $\sin x-x=-x^3/6+O(x^5)$. Therefore the limit is $-1/6$.

1. $-2$
2. $0$
3. $2$
4. $3$

Answer: (C) $2$

$f'(x)=3x^2-3=0$ gives $x=\pm1$. Since $f''(x)=6x$, $x=-1$ gives a local maximum. Therefore $f(-1)=2$.

1. $\ln(1+x^2)+C$
2. $\tan^{-1}x+C$
3. $\sin^{-1}x+C$
4. $\frac{1}{1+x^2}+C$

Answer: (B) $\tan^{-1}x+C$

The standard integral is $\int\frac{dx}{1+x^2}=\tan^{-1}x+C$.

1. $0$
2. $1/2$
3. $1$
4. $2$

Answer: (B) $1/2$

Using $\sin x\cos x=\frac12\sin2x$, the integral from $0$ to $\pi/2$ evaluates to $1/2$.

1. $y=x+C$
2. $y=Ce^x$
3. $y=Cx$
4. $y=e^{x+C^2}$

Answer: (B) $y=Ce^x$

Separating variables gives $dy/y=dx$. Integration gives $\ln y=x+C$, hence $y=Ce^x$.

1. $11/5$
2. $2$
3. $3/5$
4. $4$

Answer: (A) $11/5$

Distance $=\frac{|3(2)+4(-3)-5|}{\sqrt{3^2+4^2}}=\frac{|-11|}{5}=\frac{11}{5}$.

1. $(x-3)^2+(y-4)^2=8$
2. $(x-3)^2+(y-4)^2=4$
3. $(x-1)^2+(y-2)^2=8$
4. $(x-5)^2+(y-6)^2=8$

Answer: (A) $(x-3)^2+(y-4)^2=8$

The centre is the midpoint $(3,4)$. The diameter is $4\sqrt2$, so the radius is $2\sqrt2$ and $r^2=8$. Hence the equation is $(x-3)^2+(y-4)^2=8$.

1. $3$
2. $6$
3. $12$
4. $24$

Answer: (C) $12$

Comparing $y^2=12x$ with $y^2=4ax$, we get $4a=12$, so $a=3$. Therefore the latus rectum length is $4a=12$.

1. $1/3$
2. $1/2$
3. $2$
4. $3$

Answer: (A) $1/3$

$(\sec\theta+\tan\theta)(\sec\theta-\tan\theta)=\sec^2\theta-\tan^2\theta=1$. Therefore $\sec\theta-\tan\theta=1/3$.

1. $\pi/4$
2. $\pi/2$
3. $3\pi/4$
4. $\pi$

Answer: (B) $\pi/2$

The principal value of $\tan^{-1}(1)$ is $\pi/4$. Therefore the sum is $\pi/4+\pi/4=\pi/2$.

1. $1/6$
2. $1/9$
3. $1/12$
4. $1/18$

Answer: (B) $1/9$

Favourable pairs are $(3,6),(4,5),(5,4),(6,3)$, giving 4 outcomes out of 36. Therefore probability is $4/36=1/9$.

1. $17$
2. $50$
3. $60$
4. $72$

Answer: (C) $60$

Mean = sum/number of observations. Therefore sum $=12\times5=60$.

1. $4$
2. $6$
3. $8$
4. $10$

Answer: (C) $8$

Mean is $6$. Squared deviations are $16,4,0,4,16$, whose sum is $40$. Population variance is $40/5=8$.

1. $15$
2. $45$
3. $135$
4. $27$

Answer: (C) $135$

For an $n\times n$ matrix, $|kA|=k^n|A|$. Thus for a $3\times3$ matrix, $|3A|=3^3|A|=27\times5=135$.

1. $20$
2. $30$
3. $60$
4. $120$

Answer: (B) $30$

LEVEL contains two L's and two E's. Therefore the number of distinct arrangements is $5!/(2!2!)=30$.

1. $150$
2. $200$
3. $300$
4. $450$

Answer: (B) $200$

Choose 2 men from 5 in ${}^{5}C_2=10$ ways and 3 women from 6 in ${}^{6}C_3=20$ ways. Therefore total ways $=10\times20=200$.

1. $1/6$
2. $1/5$
3. $2/3$
4. $5/6$

Answer: (A) $1/6$

For independent events, $P(A\cap B)=P(A)P(B)=\frac12\times\frac13=\frac16$.

1. $4$
2. $6$
3. $8$
4. $10$

Answer: (C) $8$

Area $=\int_1^3 2x\,dx=[x^2]_1^3=9-1=8$.

1. $e^x$
2. $e^{-x}$
3. $x$
4. $1+x$

Answer: (A) $e^x$

For a linear differential equation $dy/dx+Py=Q$, the integrating factor is $e^{\int Pdx}$. Here $P=1$, so the integrating factor is $e^x$.