# 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.

$$-240 = 10t -\frac{1}{2}gt^2$$

$$ \therefore 5t^2-10t -240$$

$$\therefore t^2-2t-48 =0$$

$$\Rightarrow t=8,-6$$

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.

$$\displaystyle T = 2 \pi \sqrt {\frac {L} {g}} $$

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

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

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

$$\displaystyle \frac{\Delta g}{g}\times 100=\frac{1}{2}+\frac{20}{9}$$$$=2.7$$%

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

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

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$$ \dfrac{\dfrac{5}{3}mv^2}{3mv^2}\times100=56$$ %

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$$ \dfrac{\dfrac{5}{3}mv^2}{3mv^2}\times100=56$$ %

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.

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

$$I=\frac { 1 }{ 6 } M{ L }^{ 2 }\\ \Rightarrow I=\frac { 1 }{ 6 } \times \frac { 2{ M } }{ \sqrt { 3 } \pi } \times \left( \frac { 2R }{ \sqrt { 3 } } \right) ^{ 2 }=\frac { 4MR^{ 2 } }{ 9\sqrt { 3 } \pi } $$

$$\because FG=GC=L\\ \Rightarrow FC=\sqrt { { \left( FG \right) }^{ 2 }+{ \left( GC \right) }^{ 2 } } =\sqrt { { L }^{ 2 }+{ L }^{ 2 } } =\sqrt { 2 } L\\ \Rightarrow FD=\sqrt { { \left( FC \right) }^{ 2 }+{ \left( CD \right) }^{ 2 } } =\sqrt { { \left( \sqrt { 2 } L \right) }^{ 2 }+{ L }^{ 2 } } =\sqrt { 3 } L\\ \Rightarrow \sqrt { 3 } L=2R\\ \Rightarrow L=\frac { 2R }{ \sqrt { 3 } } $$

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

$$I=\frac { 1 }{ 6 } M{ L }^{ 2 }\\ \Rightarrow I=\frac { 1 }{ 6 } \times \frac { 2{ M } }{ \sqrt { 3 } \pi } \times \left( \frac { 2R }{ \sqrt { 3 } } \right) ^{ 2 }=\frac { 4MR^{ 2 } }{ 9\sqrt { 3 } \pi } $$

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) $$=-\dfrac{3GM_c}{2R_c}$$

$$V=-\dfrac{11GM}{8R}-(-\dfrac{3GM_c}{2R_c})=-\dfrac{11GM}{8R}+\dfrac{3G\dfrac{M}{8}}{2\dfrac{R}{2}}\\\Rightarrow V= -\dfrac{GM}{R}$$

Since we know that

$$T\propto \sqrt { l } \\ \Rightarrow \dfrac { { T }_{ M } }{ T } =\dfrac { \sqrt { l+\Delta l } }{ \sqrt { l } } \\ \Rightarrow { \left( \dfrac { { T }_{ M } }{ T } \right) }^{ 2 }=\dfrac { l+\Delta l }{ l } =1+\dfrac { \Delta l }{ l } \\ \Rightarrow \dfrac { \Delta l }{ l } ={ \left( \dfrac { { T }_{ M } }{ T } \right) }^{ 2 }-1\quad \quad \quad \quad .........(I)$$

$$\because Y=\dfrac{\text{stress}}{\text{strain}} \\\Rightarrow \dfrac{1}{Y}=\dfrac{\text{strain}}{\text{stress}} = \dfrac{\left (\dfrac { \Delta l }{ l } \right )}{\left ( \dfrac{Mg}{A} \right )} \\\Rightarrow \dfrac{1}{Y}=\left [\dfrac { \Delta l }{ l } \right ]\dfrac{A}{Mg} \\\Rightarrow \dfrac{1}{Y}=\left [{ \left( \dfrac { { T }_{ M } }{ T } \right) }^{ 2 }-1 \right ]\dfrac{A}{Mg} \quad \quad \text{ From} (I)$$

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

$$\Delta S^\circ{}=\int_{A}^{B}\dfrac{dQ}{T}$$...(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

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