# JEE Mains (Apr 4) 2015

For the first ball,
$-240 = 10t -\frac{1}{2}gt^2$
$\therefore 5t^2-10t -240$
$\therefore t^2-2t-48 =0$
$\Rightarrow t=8,-6$
The 1st particle will reach the ground in 8 secs.
Upto 8 secs, the relative velocity between the particles is 30m/sec and the relative acceleration is zero..
For the 2nd particle,
$-240 = 40t -5t^2$
$\Rightarrow t^2 -8t -48 =0$
$t =12\: secs$
The second particle will strike the ground in 12 secs.
Option(A): Linear behaviour after 8 secs, hence can be eliminated.
Option(B) shows increase until 12 secs , hence is not correct. after 8 secs, separation will decrease.
Option(c) is the correct one.
1
#299406
Solution
Two stones are thrown up simultaneously from the edge of a cliff 240m high with initial speed 10 m/s and 40 m/s respectively.  Which of the following graph best represents the time variation of relative position of the second stone with respect to the first ?
(Assume stones do not rebound after hitting the ground and neglect air resistance, take g = 10 m $/s^2$)
(The figures are schematic and not drawn to scale.)
A
B
C
D
$\displaystyle T = 2 \pi \sqrt {\frac {L} {g}}$

$g=4\pi^2\dfrac{L}{T^2}$

$\displaystyle \frac{\Delta g}{g}=\frac{\Delta L}{L}+2\frac{\Delta T}{T}$

$\displaystyle \frac{\Delta g}{g}=\frac{10^{-3}}{0.2}+\frac{2\times 1}{90}$

$\displaystyle \frac{\Delta g}{g}\times 100=\frac{1}{2}+\frac{20}{9}$$=2.7$%
2
#299411
Solution
The period of oscillation of a simple pendulum is $\displaystyle T = 2 \pi \sqrt {\frac {L} {g}}$. Measured of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s using a wrist watch of 1s resolution. The accuracy in the determination of g is :
A
2.5%
B
2.7%
C
2%
D
3%
As the both the masses are in equilibrium the static friction will be acting on blocks which can be found from balancing the forces in vertical direction.
The FBD of block A
Writing force balance in vertical direction for block A
$f_1=$ weight of the block A =$20$ N
Writing force balance in vertical direction for block B
$f_2-f_1=$ weight of the block B $100$ N
The value of frictional force between wall and block B is $120$ N
3
#299417
Solution
Given in the figure are two blocks A and B of weight $20$ N and $100$ N, respectively. These are being pressed against a wall by a force $F$ as shown. If the coefficient of friction between the blocks is $0.1$ and between block B and the wall is $0.15$, the frictional force applied by the wall on block B is
A
100 N
B
80 N
C
120 N
D
150 N
as the collision is inelastic the masses will move together
Assuming the speed of block A and B becomes $v_1\hat{i}+v_2\hat{j}$
Writing momentum equation in x direction
$m(2v)+2m(0)= 3m v_1 \Rightarrow v_1=\dfrac{2v}{3}$
Writing momentum equation in Y direction
$m(0)+2m(v)= 3m v_2 \Rightarrow v_2=\dfrac{2v}{3}$
The velocity of both blocks will be $\dfrac{2v}{3}\hat{i}+\dfrac{2v}{3}\hat{j}$
The loss in kinetic energy
$\dfrac{1}{2}m(2v)^2+\dfrac{1}{2}2m(v)^2-\dfrac{1}{2}3m(\dfrac{\sqrt{2}2v}{3})^2$
$3mv^2-\dfrac{4}{3}mv^2=\dfrac{5}{3}mv^2$
percentage loss$\dfrac{\dfrac{5}{3}mv^2}{3mv^2}\times100=56$ %
4
#299424
Solution
A particle of mass m moving in the direction $x$ with speed $2v$ is hit by another particle of mass 2m moving in the $y$ direction with speed $v$ . If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to
A
44%
B
50%
C
56%
D
62%
Suppose the cylindrical symmetry of the problem to note that the center of mass must lie along the z axis (x = y = 0). The only issue is how high does it lie. If the uniform density of the cone is $\rho$ , then first compute the mass of the cone. If we slice the cone into circular disks of area $\pi r^2$ and height $dz$, the mass is given by the integral:

$M = \int\limits \rho dV = \rho \int\limits_{0}^{h} \pi r^2 dz$
However, we know that the radius $r$ starts at a for $z = 0$, and goes linearly to zero when $z = h$. This means that $r = a(1 -z/h)$, so that:
$M = \rho \int\limits_{0}^{h} \pi a^2 (1 - z/h)^2 dz = \pi a^2 \rho \int\limits_{0}^{h} (1- 2z/h + z^2/h^2) dz = 1/3 \pi a^2 h \rho$
Now this simply indicates that the volume of the cone is given by

$V = \dfrac{1}{3} \pi a^2 h$
To find the height of the center of mass, we then compute: $z _{cm} = \dfrac{1}{M}\int\limits \rho z dV = \dfrac{\rho}{M} \int\limits_{0}^{h} \pi r^2 z dz = \dfrac{\pi a^2 \rho}{M} \int\limits_{0}^{h} (1- \dfrac{z}{h})^2 z dz$ $= \dfrac{\pi a^2 \rho}{M} \int\limits_{0}^{h} (z - \dfrac{2z^2}{h} + \dfrac{z^3}{h}) dz = \dfrac{1}{12M} \pi a^2 h^2 \rho = \dfrac{1}{4} h$
As a result, the center of mass of the cone is along the symmetry axis, one quarter of the way up from the base to the tip and $\dfrac{3}{4}h$ from the tip.
5
#299430
Solution
Distance of the center of mass of a solid uniform cone from its vertex is $z_0$ . If the radius of its base is R and its height is h then $z_0$ is equal to :
A
$\displaystyle \frac {h^2} {4R}$
B
$\displaystyle \frac {3h} {4}$
C
$\displaystyle \frac {5h} {8}$
D
$\displaystyle \frac {3h^2} {8R}$
When the volume of the cube is maximum, the longest diagonal of cube will be equal to diameter of the sphere.
$\because FG=GC=L\\ \Rightarrow FC=\sqrt { { \left( FG \right) }^{ 2 }+{ \left( GC \right) }^{ 2 } } =\sqrt { { L }^{ 2 }+{ L }^{ 2 } } =\sqrt { 2 } L\\ \Rightarrow FD=\sqrt { { \left( FC \right) }^{ 2 }+{ \left( CD \right) }^{ 2 } } =\sqrt { { \left( \sqrt { 2 } L \right) }^{ 2 }+{ L }^{ 2 } } =\sqrt { 3 } L\\ \Rightarrow \sqrt { 3 } L=2R\\ \Rightarrow L=\frac { 2R }{ \sqrt { 3 } }$
Since $\text{mass} \propto \text{volume}$, we have
$\frac { { M }_{ C } }{ { M }_{ S } } =\frac { { V }_{ C } }{ { V }_{ S } } \\ \Rightarrow { M }_{ C }=\frac { { V }_{ C } }{ { V }_{ S } } \times { M }_{ S }\\ \Rightarrow { M }_{ C }=\frac { { \left( \frac { 2R }{ \sqrt { 3 } } \right) }^{ 3 } }{ \frac { 4 }{ 3 } \pi { R }^{ 3 } } \times { M }\\ \Rightarrow { M }_{ C }=\frac { 2{ M } }{ \sqrt { 3 } \pi }$
And moment of inertia of cube about an axis passing through its center and perpendicular to one of its faces is given by
$I=\frac { 1 }{ 6 } M{ L }^{ 2 }\\ \Rightarrow I=\frac { 1 }{ 6 } \times \frac { 2{ M } }{ \sqrt { 3 } \pi } \times \left( \frac { 2R }{ \sqrt { 3 } } \right) ^{ 2 }=\frac { 4MR^{ 2 } }{ 9\sqrt { 3 } \pi }$
6
#299440
Solution
From a solid sphere of mass M and radius R a cube of maximum possible volume is cut. Moment of inertia of cube about an axis passing through its center and perpendicular to one of its faces is :
A
$\displaystyle \frac {MR^2} {32 \sqrt{2} \pi}$
B
$\displaystyle \frac {MR^2} {16 \sqrt{2} \pi}$
C
$\displaystyle \frac {4MR^2} {9 \sqrt{3} \pi}$
D
$\displaystyle \frac {4MR^2} {3 \sqrt{3} \pi}$
Gravitational potential at any inside point is given as
$V=-\dfrac{GM}{2R^3}(3R^2-r^2)$....(i)
for $r=\dfrac{R}{2} \;V=-\dfrac{11GM}{8R}$
Subtracting potential due to cavity of mass $M_c=\dfrac{M}{8}$ and $R_c=\dfrac{R}{2}$
Gravitational potential at center is obtained by substituting r=0 in equation (i) $=-\dfrac{3GM_c}{2R_c}$
$V=-\dfrac{11GM}{8R}-(-\dfrac{3GM_c}{2R_c})=-\dfrac{11GM}{8R}+\dfrac{3G\dfrac{M}{8}}{2\dfrac{R}{2}}\\\Rightarrow V= -\dfrac{GM}{R}$
7
#299463
Solution
From a solid sphere of mass $M$ and radius $R$, a spherical portion of radius $\displaystyle \dfrac {R} {2}$ is removed, as shown in the figure. Taking gravitational potential $V = 0$ at $r = \infty$ , the potential at the centre of the cavity thus formed is :
($G =$ gravitational constant)
A
$\displaystyle \frac {-GM} {2R}$
B
$\displaystyle \frac {-GM} {R}$
C
$\displaystyle \frac {-2GM} {3R}$
D
$\displaystyle \frac {-2GM} {R}$
Since we know that
$T\propto \sqrt { l } \\ \Rightarrow \dfrac { { T }_{ M } }{ T } =\dfrac { \sqrt { l+\Delta l } }{ \sqrt { l } } \\ \Rightarrow { \left( \dfrac { { T }_{ M } }{ T } \right) }^{ 2 }=\dfrac { l+\Delta l }{ l } =1+\dfrac { \Delta l }{ l } \\ \Rightarrow \dfrac { \Delta l }{ l } ={ \left( \dfrac { { T }_{ M } }{ T } \right) }^{ 2 }-1\quad \quad \quad \quad .........(I)$
$\because Y=\dfrac{\text{stress}}{\text{strain}} \\\Rightarrow \dfrac{1}{Y}=\dfrac{\text{strain}}{\text{stress}} = \dfrac{\left (\dfrac { \Delta l }{ l } \right )}{\left ( \dfrac{Mg}{A} \right )} \\\Rightarrow \dfrac{1}{Y}=\left [\dfrac { \Delta l }{ l } \right ]\dfrac{A}{Mg} \\\Rightarrow \dfrac{1}{Y}=\left [{ \left( \dfrac { { T }_{ M } }{ T } \right) }^{ 2 }-1 \right ]\dfrac{A}{Mg} \quad \quad \text{ From} (I)$
8
#299492
Solution
A pendulum of a uniform wire of cross sectional area A has time period $T.$ When an additional mass $M$ is added to its bob, the time period changes to $T_M$ . If the Young's modulus of the material of the wire is $Y$ then $\displaystyle \frac {1} {Y}$ is equal to :
($G$ = gravitational acceleration)
A
$\displaystyle \left[ \left( \frac {T_M} {T} \right)^2 - 1 \right] \frac {A} {M_g}$
B
$\displaystyle \left[ \left( \frac {T_M} {T} \right)^2 - 1 \right] \frac {M_g} {A}$
C
$\displaystyle \left[ 1 - \left( \frac {T_M} {T} \right)^2 \right] \frac {A} {M_g}$
D
$\displaystyle \left[ 1 - \left( \frac {T} {T_M} \right)^2 \right] \frac {A} {M_g}$
9
#299516
Consider a spherical shell of radius $R$ at temperature $T$. The black body radiation inside it can be considered as an ideal gas of photons with internal energy per unit volume  $u = \displaystyle \dfrac {U} {V} \propto T^4$ and pressure  $P = \displaystyle \dfrac {1} {3} \left( \frac {U} {V} \right)$ . If the shell now undergoes an adiabatic expansion the relation between T and R is :
A
$T \propto e^{-R}$
B
$T \propto e^{-3R}$
C
$\displaystyle T \propto \dfrac {1} {R}$
D
$\displaystyle T \propto \dfrac {1} {R^3}$
The entropy change from State A to B is given by
$\Delta S^\circ{}=\int_{A}^{B}\dfrac{dQ}{T}$...(i)
where $dQ$ is heat supplied in J to the body and temperature is in K

$T_A=273+100 K=373 K;T_B=273+200=473 K$
$dQ=CdT$ where C is heat capacity of body and dT is increase in temperature in $C^\circ{}$

Substituting $T'=T-273$ in equation (i) where $T'$ is temperature in $C^\circ{}$ and $T$ is in kelvin.
$dT'=dT$
integration limits:$T'_A=373-273=100 C^\circ{}\;;T'_B=473-273 =200C^\circ{}$
$\Delta S^\circ{}=\int_{100}^{200}\dfrac{CdT'}{T'+273}=ln(\dfrac{473}{373})$
As the heat transferred is same the entropy change will be same in both the cases
10
#299541
Solution
A solid body of constant heat capacity $1 J/ ^{\circ}C$ is being heated by keeping it in contact with reservoirs in two ways :
(i) Sequentially keeping in contact with $2$ reservoirs such that each reservoir supplies same amount of heat.
(ii) Sequentially keeping in contact with $8$ reservoirs such that each reservoir supplies same amount of heat.
In both the cases body is brought from initial temperature of $100^{\circ}C$ to a final temperature of $200^{\circ}C$ . Entropy change of the body in two cases respectively is :
A
$\ln2 , 4\ln2$
$\ln2 , \ln2$
$\ln2 , 2\ln2$
$2\ln2 , 8\ln2$