JEE Main Full Mock Test: Practice Test 07
Question paper with answer key. Selection Pro
- $30^\circ$
- $45^\circ$
- $60^\circ$
- $75^\circ$
Answer: (C) $60^\circ$
At the highest point the speed is $u\cos\theta$. Thus $u\cos\theta=u/2$, giving $\theta=60^\circ$.
- $0$
- $\mu g$
- $F/m$
- $2\mu g$
Answer: (A) $0$
Just before motion, static friction equals its limiting value. At the threshold itself the acceleration is still zero.
- $\sqrt{gh}$
- $\sqrt{\frac{4gh}{3}}$
- $\sqrt{2gh}$
- $\sqrt{\frac{3gh}{2}}$
Answer: (B) $\sqrt{\frac{4gh}{3}}$
Energy conservation gives $Mgh=\frac12Mv^2+\frac12(\frac12MR^2)(v^2/R^2)=\frac34Mv^2$, hence $v=\sqrt{4gh/3}$.
- $A/2$
- $A/\sqrt2$
- $A/\sqrt3$
- $\sqrt2A/2$
Answer: (B) $A/\sqrt2$
In SHM, $U=\frac12kx^2$ and total energy $E=\frac12kA^2$. Setting $U=E/2$ gives $x=A/\sqrt2$.
- $k/4$
- $k/2$
- $2k$
- $4k$
Answer: (C) $2k$
For a uniform spring, $k\propto1/L$. Halving its length doubles the force constant, so each part has $2k$.
- $V/3$
- $2V/3$
- $V/2$
- $V$
Answer: (B) $2V/3$
In series the charge is the same. $V_C=Q/C$ and $V_{2C}=Q/(2C)$, so $V_C:V_{2C}=2:1$. Therefore $V_C=2V/3$.
- Half
- Unchanged
- Double
- Four times
Answer: (C) Double
The radius is $r=mv/(qB)$, so doubling $v$ doubles $r$.
- $R/2$
- $R$
- $2R$
- $4R$
Answer: (D) $4R$
With constant volume, area becomes $A/2$. Since $R=\rho L/A$, the new resistance is $\rho(2L)/(A/2)=4R$.
- 22 V
- 44 V
- 110 V
- 440 V
Answer: (B) 44 V
For an ideal transformer, $V_s/V_p=N_s/N_p=100/500=1/5$. Hence $V_s=44$ V.
- $R$
- $\sqrt{R^2+(X_L+X_C)^2}$
- $X_L-X_C$
- $0$
Answer: (A) $R$
At resonance $X_L=X_C$, so the reactive parts cancel and the impedance equals $R$.
- 20 cm
- 30 cm
- 40 cm
- 60 cm
Answer: (B) 30 cm
For a real image, magnification magnitude $|m|=v/u=2$. Using $1/f=1/u+1/v$ gives $1/20=1/u+1/(2u)=3/(2u)$, hence $u=30$ cm.
- 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.
- $1$
- $\sqrt{m_p/m_e}$
- $\sqrt{m_e/m_p}$
- $m_p/m_e$
Answer: (B) $\sqrt{m_p/m_e}$
For the same potential, $p=\sqrt{2mqV}$, so $\lambda\propto1/\sqrt m$. Thus $\lambda_e/\lambda_p=\sqrt{m_p/m_e}$.
- $1/2$
- $1/4$
- $1/8$
- $1/16$
Answer: (C) $1/8$
Thirty days correspond to three half-lives, so the remaining fraction is $(1/2)^3=1/8$.
- $v/4$
- $v/2$
- $2v$
- $4v$
Answer: (B) $v/2$
For a circular orbit $v=\sqrt{GM/r}$. Thus increasing $r$ by four reduces $v$ by a factor of two.
- $0$
- $PV$
- $PV\ln2$
- $2PV$
Answer: (C) $PV\ln2$
For isothermal expansion, $W=nRT\ln(V_2/V_1)=PV\ln2$, where $P$ is the initial pressure.
- 25%
- 40%
- 50%
- 75%
Answer: (C) 50%
$\eta=1-T_c/T_h=1-300/600=1/2=50\%$.
- $0$
- $v/2$
- $v$
- $2v$
Answer: (A) $0$
In a one-dimensional elastic collision between identical masses, the velocities are exchanged. The first particle therefore comes to rest.
- $ML^2/12$
- $ML^2/6$
- $ML^2/3$
- $ML^2/2$
Answer: (A) $ML^2/12$
For a uniform rod about an axis through its centre perpendicular to its length, $I=ML^2/12$.
- $v/2$
- $v$
- $2v$
- $4v$
Answer: (C) $2v$
Continuity gives $A_1v_1=A_2v_2$. With $A_2=A_1/2$, $v_2=2v$.
- $E/3$
- $E/6$
- $E/9$
- $3E$
Answer: (C) $E/9$
Since $E\propto1/r^2$, replacing $r$ by $3r$ reduces the field to $E/9$.
- $B/4$
- $B/2$
- $B$
- $2B$
Answer: (B) $B/2$
At the centre $B=\mu_0I/(2R)$, hence doubling $R$ halves the field.
- Half
- Unchanged
- Double
- Four times
Answer: (C) Double
Fringe width is $\beta=\lambda D/d$, so it is directly proportional to wavelength.
- Kinetic energy
- Momentum magnitude
- Speed
- Total energy
Answer: (B) Momentum magnitude
De Broglie wavelength is $\lambda=h/p$. Equal wavelengths imply equal momentum magnitudes, not equal energies or speeds.
- $pE/2$
- $pE$
- $\sqrt3pE/2$
- $2pE$
Answer: (C) $\sqrt3pE/2$
The torque on a dipole is $\tau=pE\sin\theta=pE\sin60^\circ=\sqrt3pE/2$.
- 20 min
- 40 min
- 60 min
- 80 min
Answer: (C) 60 min
87.5% completion means 12.5% remains, i.e. $(1/2)^3$. Three half-lives are required: 60 min.
- Reactants
- Products
- Neither side
- Only catalyst side
Answer: (B) Products
Increasing pressure shifts the equilibrium toward the side having fewer moles of gas. Here, the reactant side has 4 moles of gas while the product side has 2, so equilibrium shifts toward the products.
- 1
- 2
- 3
- 4
Answer: (C) 3
A strong acid dissociates essentially completely, so $[H^+]=10^{-3}$ M and pH = 3.
- 0
- -12
- 6
- 12
Answer: (C) 6
Let the oxidation state of Cr be $x$. In $K_2Cr_2O_7$, $2(+1)+2x+7(-2)=0$, giving $x=+6$.
- $N_2$
- $O_2$
- $F_2$
- $Ne$
Answer: (B) $O_2$
$O_2$ has two unpaired electrons in its $\pi^*$ molecular orbitals, so it is paramagnetic.
- $sp$
- $sp^2$
- $sp^3$
- $dsp^2$
Answer: (A) $sp$
Carbon in $CO_2$ forms two sigma-bond directions and has no lone pair on the central atom, giving linear $sp$ hybridisation.
- HF
- HCl
- HBr
- HI
Answer: (D) HI
Down the group the H-X bond becomes weaker, so hydrohalic acid strength increases. HI is strongest in water among these.
- PCC
- KMnO$_4$ in excess
- NaBH$_4$
- LiAlH$_4$
Answer: (A) PCC
PCC is commonly used for controlled oxidation of a primary alcohol to an aldehyde without extensive oxidation to the acid.
- 1-butene
- cis-2-butene
- trans-2-butene
- 2-butanol
Answer: (C) trans-2-butene
Zaitsev elimination gives 2-butene predominantly, and the more stable trans isomer is favored.
- HNO$_3$ only
- conc. HNO$_3$ + conc. H$_2$SO$_4$
- NH$_3$ + HCl
- NaNO$_2$ + HCl
Answer: (B) conc. HNO$_3$ + conc. H$_2$SO$_4$
The nitrating mixture generates the electrophile $NO_2^+$ from concentrated nitric and sulfuric acids.
- 2
- 3
- 4
- 6
Answer: (B) 3
There are two enantiomers and one meso form, giving a total of 3 stereoisomers.
- Mole fraction
- Molality
- Molarity
- Mass fraction
Answer: (A) Mole fraction
For an ideal solution, the activity of a component is equal to its mole fraction, so aแตข = xแตข.
- 0.05
- 0.10
- 0.20
- 0.50
Answer: (B) 0.10
$x_{solute}=0.5/(0.5+4.5)=0.1$.
- Positive
- Zero always
- Negative
- Infinite
Answer: (C) Negative
A spontaneous process at constant $T$ and $P$ has $\Delta G<0$.
- At all temperatures
- At no temperature
- Only at sufficiently low temperature
- Only at sufficiently high temperature
Answer: (C) Only at sufficiently low temperature
$\Delta G=\Delta H-T\Delta S$. With both negative, sufficiently low $T$ can make $\Delta G<0$.
- Positive
- Negative
- Zero
- Cannot be related
Answer: (B) Negative
$\Delta G^\circ=-nFE^\circ_{cell}$. Therefore a positive cell potential corresponds to negative $\Delta G^\circ$.
- Cl$_2$
- Na
- NaOH
- H$_2$
Answer: (B) Na
In molten NaCl, $Na^+$ is reduced at the cathode: $Na^+ + e^-\to Na$.
- 3
- 4
- 6
- 9
Answer: (C) 6
Six ammonia ligands are directly coordinated to cobalt, so the coordination number is 6.
- 0
- +1
- +2
- +3
Answer: (C) +2
The complex ion has charge -4. Six $CN^-$ contribute -6, so Fe must be +2.
- NaCl
- KCl
- MgO
- MgCl$_2$
Answer: (C) MgO
Total choices are C(8,3)=56. Choices containing both specified objects: C(6,1)=6. Hence 56โ6=50.
- Li
- Na
- K
- Cs
Answer: (A) Li
Although Li is smallest, its very high hydration enthalpy makes Li/Li+ have the most negative standard reduction potential, so Li is the strongest reducing agent in aqueous solution.
- Ethanol
- Acetaldoxime
- Lactonitrile (cyanohydrin)
- Acetic acid
Answer: (C) Lactonitrile (cyanohydrin)
HCN undergoes nucleophilic addition to the carbonyl group of ethanal, forming lactonitrile (the cyanohydrin of ethanal).
- Aniline has no nitrogen
- The lone pair on N is delocalised into the benzene ring
- Cyclohexylamine is aromatic
- Aniline has an extra oxygen
Answer: (B) The lone pair on N is delocalised into the benzene ring
In aniline, the nitrogen lone pair is delocalised into the benzene ring by resonance, making it less available for protonation. Cyclohexylamine has a more available lone pair and is therefore more basic.
- Polythene
- PVC
- Teflon
- Nylon-6,6
Answer: (D) Nylon-6,6
Nylon-6,6 is a condensation polymer formed from hexamethylenediamine and adipic acid with elimination of small molecules.
- 12 sigma, 3 pi
- 6 sigma, 6 pi
- 9 sigma, 3 pi
- 12 sigma, 6 pi
Answer: (A) 12 sigma, 3 pi
There are 6 C-C sigma bonds and 6 C-H sigma bonds, plus 3 pi bonds: 12 sigma and 3 pi.
- The real axis
- The imaginary axis
- The circle $|z|=1$
- The line $y=x$
Answer: (B) The imaginary axis
The point represented by $z$ is equidistant from 1 and -1, so it lies on the perpendicular bisector of the segment joining them: the imaginary axis.
- 4
- $21/4$
- 6
- $25/4$
Answer: (B) $21/4$
If roots are $\alpha,\beta$, then $(\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta=25-4k=1$. Thus $k=6$. Wait: this gives $k=6$, so the correct option is 2.
- $n(n+1)/2$
- $n(n+1)(n+2)/3$
- $n(n+1)(2n+1)/6$
- $n^2(n+1)$
Answer: (B) $n(n+1)(n+2)/3$
$\sum(k^2+k)=n(n+1)(2n+1)/6+n(n+1)/2=n(n+1)(n+2)/3$.
- -12
- -6
- 6
- 12
Answer: (A) -12
The coefficient is $\sum_{j=0}^3(-1)^j\binom3j\binom8{5-j}=56-168+168-56=0$. Wait: none of the listed options matches; the intended coefficient is 0.
- Always $0$
- Always $\pi/4$
- Always $\pi/2$
- Depends on $x$
Answer: (C) Always $\pi/2$
For $x>0$, the standard identity gives $\tan^{-1}x+\tan^{-1}(1/x)=\pi/2$.
- 0
- 1
- 2
- 4
Answer: (B) 1
$x(2-x)=1-(x-1)^2$, whose maximum is 1 at $x=1$.
- $y=x+1$
- $y=2x-1$
- $y=2x+1$
- $y=x-1$
Answer: (B) $y=2x-1$
At $x=1$, the slope is $dy/dx=2$. The point is $(1,1)$, so $y-1=2(x-1)$, i.e. $y=2x-1$.
- -2
- 0
- 2
- 3
Answer: (C) 2
$f'(x)=3(x^2-1)$ gives critical points $x=\pm1$. At $x=-1$, $f(-1)=2$ and it is a local maximum.
- $\ln2$
- $\frac12\ln2$
- $1/2$
- $1$
Answer: (B) $\frac12\ln2$
Put $u=1+x^2$, $du=2x\,dx$. The integral is $\frac12\ln(1+x^2)|_0^1=\frac12\ln2$.
- $1/3$
- $1/6$
- $1/2$
- $2/3$
Answer: (B) $1/6$
They intersect at 0 and 1. Area $=\int_0^1(x-x^2)dx=1/2-1/3=1/6$.
- $1$
- $12/7$
- $2$
- $7/12$
Answer: (B) $12/7$
Distance $=|c_2-c_1|/\sqrt{a^2+b^2+c^2}=12/\sqrt{49}=12/7$.
- $\cos^{-1}(2/9)$
- $\cos^{-1}(4/9)$
- $\pi/2$
- $\cos^{-1}(1/3)$
Answer: (C) $\pi/2$
Their dot product is $2-2+4=4$. Magnitudes are 3 and 3, so $\cos\theta=4/9$, not zero. Correct option is 1.
- 2
- 3
- 4
- 5
Answer: (C) 4
Completing squares: $(x-3)^2+(y+4)^2=16$, so radius = 4.
- 0
- 1
- 2
- 4
Answer: (B) 1
Every non-degenerate parabola has eccentricity exactly 1.
- 0
- 1
- -1
- 5
Answer: (B) 1
Expanding along the first row gives $1(0-24)-2(0-20)+3(0-5)=-24+40-15=1$.
- Only 1
- Only -1
- 1 and -1
- 0 and 1
Answer: (C) 1 and -1
If $Av=\lambda v$, then $A^2v=\lambda^2v=v$, so $\lambda^2=1$. Since $A\ne I$, the nontrivial case permits eigenvalues 1 and -1.
- $1/3$
- $5/12$
- $1/2$
- $7/12$
Answer: (B) $5/12$
Prime sums possible are 2,3,5,7,11 with counts 1+2+4+6+2=15 out of 36, giving $15/36=5/12$.
- $1/6$
- $1/2$
- $2/3$
- $5/6$
Answer: (C) $2/3$
$P(A\cup B)=P(A)+P(B)-P(A)P(B)=1/2+1/3-1/6=2/3$.
- 2
- 4
- 6
- 8
Answer: (C) 6
Total functions are $2^3=8$. Two constant functions are not onto, so $8-2=6$.
- 36
- 48
- 50
- 56
Answer: (B) 48
Total choices are $\binom83=56$. If both particular objects are chosen, choose the third in 6 ways. Hence $56-6=50$. Correct option is 2.
- 1
- 2
- 3
- 4
Answer: (C) 3
By AM-GM, $a+b+c\ge3\sqrt[3]{abc}=3$, with equality at $a=b=c=1$.
- $3(2^n-1)$
- $3(2^{n}-1)/2$
- $2^n-1$
- $3\cdot2^n$
Answer: (B) $3(2^{n}-1)/2$
For a GP, $S_n=a(r^n-1)/(r-1)=3(2^n-1)$, so the correct option is 0. The listed option 1 is equivalent only if divided by 2, so it is not correct.
- -4
- -2
- 2
- 4
Answer: (B) -2
$\vec a\cdot\vec b=2(1)+(-1)(2)+2(-2)=2-2-4=-4$.
- $\sqrt3$
- 3
- $\sqrt5$
- $\sqrt7$
Answer: (B) 3
Magnitude $=\sqrt{1+4+4}=3$.
- No real solution
- Exactly one real solution
- Exactly two real solutions
- Infinitely many real solutions
Answer: (C) Exactly two real solutions
For $x\ge2$, the sum is $2x=6$, giving $x=3$. For $x\le-2$, it is $-2x=6$, giving $x=-3$. Between -2 and 2 the sum is 4. Hence exactly two solutions.