JEE Main Full Mock Test: Practice Test 05
Question paper with answer key. Selection Pro
- $0.5\,\text{m/s}$
- $1\,\text{m/s}$
- $\sqrt{2}\,\text{m/s}$
- $2\,\text{m/s}$
Answer: (B) $1\,\text{m/s}$
For SHM, $\frac{1}{2}kA^2=\frac{1}{2}mv_{\max}^2$. Hence $v_{\max}=A\sqrt{\frac{k}{m}}=0.10\sqrt{100}=1\,\text{m/s}$.
- $10\sqrt{3}\,\text{m}$
- $20\sqrt{3}\,\text{m}$
- $30\,\text{m}$
- $40\,\text{m}$
Answer: (B) $20\sqrt{3}\,\text{m}$
$R=\frac{u^2\sin2\theta}{g}=\frac{400\sin60^\circ}{10}=20\sqrt{3}\,\text{m}$.
- $\frac{1}{8}mu^2$
- $\frac{1}{4}mu^2$
- $\frac{1}{2}mu^2$
- $\frac{3}{4}mu^2$
Answer: (B) $\frac{1}{4}mu^2$
Common speed is $u/2$. Initial KE is $\frac12mu^2$ and final KE is $\frac12(2m)(u/2)^2=\frac14mu^2$. Loss is $\frac14mu^2$.
- $v/2$
- $v/4$
- $2v$
- $4v$
Answer: (A) $v/2$
Orbital speed varies as $v\propto 1/\sqrt{r}$. Thus $v'=v/\sqrt4=v/2$.
- $4\pi\varepsilon_0RV^2$
- $2\pi\varepsilon_0RV^2$
- $\frac12(4\pi\varepsilon_0R)V^2$
- $\frac{V^2}{4\pi\varepsilon_0R}$
Answer: (C) $\frac12(4\pi\varepsilon_0R)V^2$
Capacitance of an isolated sphere is $C=4\pi\varepsilon_0R$. Hence $U=\frac12CV^2=2\pi\varepsilon_0RV^2$.
- $V/3$
- $2V/3$
- $V/2$
- $V$
Answer: (B) $2V/3$
Series capacitors carry equal charge. $C_{eq}=\frac{2C}{3}$, so $Q=\frac{2CV}{3}$. Voltage across $C$ is $Q/C=2V/3$.
- $E+Ir$
- $E-Ir$
- $IR+Ir+E$
- $E/R$
Answer: (B) $E-Ir$
For a discharging cell, terminal voltage is $V=E-Ir=IR$.
- Half
- Unchanged
- Double
- Four times
Answer: (C) Double
$F=BIL$ for perpendicular geometry, so doubling $L$ doubles the force.
- $r/2$
- $r$
- $2r$
- $4r$
Answer: (C) $2r$
The radius is $r=mv/(eB)$, hence it is directly proportional to $v$.
- $NBA$
- $NBA\omega$
- $NBA/\omega$
- $B/NA\omega$
Answer: (B) $NBA\omega$
For rotating coil, $\mathcal{E}_{\max}=NBA\omega$.
- $110\,\text{V}$
- $220\,\text{V}$
- $220\sqrt2\,\text{V}$
- $440\,\text{V}$
Answer: (C) $220\sqrt2\,\text{V}$
$V_0=\sqrt2V_{rms}=220\sqrt2\,\text{V}$.
- $10\,\text{cm}$
- $30\,\text{cm}$
- $-30\,\text{cm}$
- $-10\,\text{cm}$
Answer: (B) $30\,\text{cm}$
$\frac1F=\frac1{15}-\frac1{30}=\frac1{30}$, hence $F=30\,\text{cm}$.
- Half
- Unchanged
- Double
- Four times
Answer: (C) Double
Fringe width is $\beta=\lambda D/d$, so it doubles with wavelength.
- $100\,\text{nm}$
- $200\,\text{nm}$
- $400\,\text{nm}$
- $620\,\text{nm}$
Answer: (B) $200\,\text{nm}$
Using $\lambda(\text{nm})\approx1240/E(\text{eV})$, $\lambda\approx1240/6.2=200\,\text{nm}$.
- $2\,\text{eV}$
- $3\,\text{eV}$
- $5\,\text{eV}$
- $7\,\text{eV}$
Answer: (B) $3\,\text{eV}$
Einstein's equation gives $K_{\max}=E_\gamma-\phi=5-2=3\,\text{eV}$.
- $1/2$
- $1/4$
- $1/8$
- $1/16$
Answer: (C) $1/8$
$12$ hours corresponds to 3 half-lives, so remaining fraction is $(1/2)^3=1/8$.
- $3\lambda$
- $\lambda$
- $\lambda/3$
- $\lambda/9$
Answer: (C) $\lambda/3$
Since $\lambda=h/p$, tripling momentum makes wavelength one-third.
- $300\,\text{J}$
- $700\,\text{J}$
- $-300\,\text{J}$
- $-700\,\text{J}$
Answer: (A) $300\,\text{J}$
First law: $\Delta U=Q-W=500-200=300\,\text{J}$.
- $RT\ln3$
- $3RT\ln3$
- $RT/\ln3$
- $2RT$
Answer: (A) $RT\ln3$
For one mole in isothermal expansion, $W=RT\ln(V_2/V_1)=RT\ln3$.
- $YA/L$
- $YL/A$
- $Y/(AL)$
- $A/(YL)$
Answer: (A) $YA/L$
Stress $=Y\times$ strain gives $F/\Delta L=YA/L$.
- $T/R$
- $2T/R$
- $3T/R$
- $4T/R$
Answer: (B) $2T/R$
For a liquid drop with one surface, $\Delta P=2T/R$.
- $g\sin\theta-\mu g\cos\theta$
- $-g(\sin\theta+\mu\cos\theta)$
- $g(\sin\theta+\mu\cos\theta)$
- $-g(\sin\theta-\mu\cos\theta)$
Answer: (B) $-g(\sin\theta+\mu\cos\theta)$
Both gravity component and friction act down the incline, so acceleration is $-g(\sin\theta+\mu\cos\theta)$.
- $\omega L$
- $\omega L/2$
- $2\omega L$
- $\omega/L$
Answer: (B) $\omega L/2$
Linear speed is $v=\omega r$; at midpoint $r=L/2$.
- Disc
- Ring
- Both equal
- Cannot be determined
Answer: (B) Ring
$K=\frac12Mv^2+\frac12I\omega^2$. The ring has $I=MR^2$, larger than the disc's $MR^2/2$.
- $T/9$
- $T/3$
- $3T$
- $9T$
Answer: (C) $3T$
$T\propto\sqrt L$, so $T'=\sqrt9T=3T$.
- $1$
- $2$
- $3$
- $11$
Answer: (C) $3$
For a strong acid, $[H^+]=10^{-3}\,\text{M}$, so pH $=3$.
- $20$ min
- $40$ min
- $60$ min
- $80$ min
Answer: (C) $60$ min
Remaining fraction is $12.5\%=1/8=(1/2)^3$, so three half-lives are required: $60$ min.
- Reactants
- Products
- No change
- Both equally
Answer: (B) Products
There are 4 moles of gaseous reactants and 2 moles of gaseous products, so higher pressure favors products.
- $+4$
- $+5$
- $+6$
- $-2$
Answer: (C) $+6$
$2(+1)+x+4(-2)=0$, giving $x=+6$.
- $N_2$
- $O_2$
- $CO$
- $CN^-$
Answer: (B) $O_2$
Molecular orbital theory shows $O_2$ has two unpaired electrons in $\pi^*$ orbitals.
- $sp^2$
- $sp^3$
- $sp^3d$
- $sp^3d^2$
Answer: (D) $sp^3d^2$
Six sigma bonds around sulfur correspond to octahedral arrangement and $sp^3d^2$ hybridisation.
- Most negative $E^\circ$
- Most positive $E^\circ$
- Zero $E^\circ$
- Largest atomic mass
Answer: (B) Most positive $E^\circ$
A more positive standard reduction potential indicates greater tendency to undergo reduction and hence stronger oxidising character.
- $0.1\,M$
- $0.2\,M$
- $0.4\,M$
- $0.8\,M$
Answer: (C) $0.4\,M$
Moles of NaOH $=4/40=0.1$. Volume $=0.5$ L, so molarity $=0.2\,M$.
- $NH_3$
- $H_2O$
- $CH_4$
- $BF_3$
Answer: (D) $BF_3$
$BF_3$ is trigonal planar with bond angle $120^\circ$, larger than the others listed.
- 1-bromopropane
- 2-bromopropane
- 1,2-dibromopropane
- Propane
Answer: (B) 2-bromopropane
In the absence of peroxide, HBr adds by Markovnikov's rule to give 2-bromopropane.
- Methanol
- Ethanol
- Propanol
- Methanoic acid
Answer: (B) Ethanol
Ethanol is oxidised to ethanal, which contains the required $CH_3CO-$ group for the iodoform reaction.
- $2$
- $3$
- $4$
- $6$
Answer: (B) $3$
It has two stereocentres and a meso form, giving three stereoisomers in total.
- $H_2/Ni$
- $Br_2/CCl_4$
- $KMnO_4$
- $O_3/Zn$
Answer: (B) $Br_2/CCl_4$
Bromine adds across the double bond to form a vicinal dibromide.
- 2-methyl-1-butene
- 2-methyl-2-butene
- 1-pentene
- 2-pentene
Answer: (B) 2-methyl-2-butene
Dehydration follows Zaitsev's rule; the more substituted alkene, 2-methyl-2-butene, predominates.
- $I^-$
- $Br^-$
- $F^-$
- $CN^-$
Answer: (D) $CN^-$
$CN^-$ is a strong-field ligand and causes significant splitting in coordination complexes.
- $4$
- $5$
- $6$
- $7$
Answer: (C) $6$
There are five $NH_3$ ligands and one $Cl^-$ ligand directly attached to cobalt, giving coordination number $6$.
- Tetrahedral
- Square planar
- Octahedral
- Trigonal planar
Answer: (B) Square planar
$CN^-$ is strong field and $Ni^{2+}$ in this complex forms a square-planar $dsp^2$ arrangement.
- Nitrobenzene
- Aniline
- Phenol
- Benzoic acid
Answer: (A) Nitrobenzene
Electrophilic substitution of benzene with nitrating mixture gives nitrobenzene.
- $-OH$
- $-NH_2$
- $-CH_3$
- $-NO_2$
Answer: (D) $-NO_2$
The strongly electron-withdrawing $-NO_2$ group is deactivating and meta-directing.
- Phenol
- Aniline
- Azobenzene
- Benzene
Answer: (B) Aniline
$Sn/HCl$ reduces the nitro group to an amino group, giving aniline.
- Ethene
- Vinyl chloride
- Styrene
- Tetrafluoroethene
Answer: (B) Vinyl chloride
PVC is polyvinyl chloride, formed by polymerisation of $CH_2=CHCl$.
- $2$
- $3$
- $4$
- $5$
Answer: (C) $4$
There are three sigma bonds: one C-C and two C-H bonds.
- $0$
- $1$
- $RT$
- $P/V$
Answer: (B) $1$
For an ideal gas, $PV=nRT$, so $Z=PV/(nRT)=1$.
- Bond enthalpy
- Standard enthalpy of formation
- Lattice enthalpy
- Ionisation enthalpy
Answer: (B) Standard enthalpy of formation
This is the definition of standard enthalpy of formation.
- $CH_4$
- $NH_3$
- $H_2O$
- $H_2S$
Answer: (C) $H_2O$
Water forms an extensive three-dimensional hydrogen-bond network, giving it an unusually high boiling point.
- $5$
- $7$
- $25$
- $49$
Answer: (C) $25$
$|z|^2=3^2+4^2=25$.
- $28$
- $56$
- $70$
- $84$
Answer: (B) $56$
Coefficient is ${}^{8}C_3=56$.
- $6$
- $8$
- $9$
- $12$
Answer: (C) $9$
Equal roots require discriminant zero: $36-4k=0$, hence $k=9$.
- $45$
- $50$
- $55$
- $60$
Answer: (C) $55$
$\sum r=n(n+1)/2=10\times11/2=55$.
- $3$
- $5$
- $8$
- $11$
Answer: (B) $5$
Determinant $=2(4)-1(3)=5$.
- $0^\circ$
- $45^\circ$
- $90^\circ$
- $180^\circ$
Answer: (C) $90^\circ$
$\vec a\cdot\vec b=|a||b|\cos\theta=0$ and both vectors are non-zero, so $\theta=90^\circ$.
- $y=x+1$
- $y=2x-1$
- $y=2x+1$
- $y=x-1$
Answer: (B) $y=2x-1$
At $(1,1)$, slope is $2$. Hence $y-1=2(x-1)$, or $y=2x-1$.
- $2$
- $4$
- $6$
- $8$
Answer: (B) $4$
$x(4-x)=4-(x-2)^2$, so maximum value is $4$.
- $1/2$
- $1$
- $2$
- $4$
Answer: (B) $1$
$\int_0^1 2x\,dx=[x^2]_0^1=1$.
- $x^x$
- $x^x\ln x$
- $x^x(\ln x+1)$
- $x^{x-1}$
Answer: (C) $x^x(\ln x+1)$
Taking logarithms, $\ln y=x\ln x$ gives $y'/y=\ln x+1$, hence $y'=x^x(\ln x+1)$.
- $0$
- $1$
- $e$
- $\infty$
Answer: (B) $1$
This is the standard exponential limit and equals $1$.
- $4$
- $5$
- $6$
- $7$
Answer: (B) $5$
Distance $=\sqrt{(4-1)^2+(6-2)^2}=\sqrt{25}=5$.
- $(x-2)^2+(y+1)^2=9$
- $(x+2)^2+(y-1)^2=9$
- $(x-2)^2+(y-1)^2=3$
- $(x+2)^2+(y+1)^2=3$
Answer: (A) $(x-2)^2+(y+1)^2=9$
Circle equation is $(x-h)^2+(y-k)^2=r^2$.
- $0$
- $1$
- $1/2$
- $\sqrt2$
Answer: (B) $1$
Every parabola has eccentricity $e=1$.
- Less than $1$
- Equal to $1$
- Greater than $1$
- Zero
Answer: (C) Greater than $1$
A hyperbola has $e>1$.
- $\pi/6$
- $\pi/4$
- $\pi/3$
- $\pi/2$
Answer: (B) $\pi/4$
In the first quadrant, $\tan\theta=1$ gives $\theta=\pi/4$.
- $0$
- $\pi/4$
- $\pi/2$
- $\pi$
Answer: (D) $\pi$
The principal range of $\cos^{-1}x$ is $[0,\pi]$, so $\cos^{-1}(-1)=\pi$.
- $1/4$
- $1/2$
- $3/4$
- $1$
Answer: (B) $1/2$
Favourable outcomes are HT and TH, so probability is $2/4=1/2$.
- $6$
- $7$
- $8$
- $9$
Answer: (B) $7$
Mean $=(3+5+7+9+11)/5=35/5=7$.
- $0$
- $1$
- $2$
- $4$
Answer: (A) $0$
All observations are equal, so every deviation from the mean is zero and variance is $0$.
- $4$
- $6$
- $8$
- $16$
Answer: (C) $8$
For an $n\times n$ matrix, $|kA|=k^n|A|$. Thus $|2A|=2^3\times2=16$.
- $y=x^3+C$
- $y=3x^3+C$
- $y=x^2+C$
- $y=6x+C$
Answer: (A) $y=x^3+C$
Integrating gives $y=\int3x^2dx=x^3+C$.
- $1$
- $2$
- $3$
- $4$
Answer: (B) $2$
Area $=\int_0^2x\,dx=[x^2/2]_0^2=2$.
- $7$
- $11$
- $13$
- $25$
Answer: (C) $13$
$\alpha+\beta=5$ and $\alpha\beta=6$. Hence $\alpha^2+\beta^2=25-12=13$.
- $24$
- $48$
- $56$
- $64$
Answer: (C) $56$
Number of choices is ${}^{8}C_3=56$.