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.
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#299406
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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
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B
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C
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D
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$$\displaystyle T = 2 \pi \sqrt {\frac {L} {g}} $$

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

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

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

$$\displaystyle \frac{\Delta g}{g}\times 100=\frac{1}{2}+\frac{20}{9}$$$$=2.7$$%
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#299411
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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%
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B
2.7%
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C
2%
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D
3%
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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
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#299417
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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
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B
80 N
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C
120 N
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D
150 N
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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$$ \dfrac{\dfrac{5}{3}mv^2}{3mv^2}\times100=56$$ %
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#299424
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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%
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B
50%
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C
56%
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D
62%
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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.
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#299430
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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
$$\displaystyle \frac {h^2} {4R} $$
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B
$$\displaystyle \frac {3h} {4} $$
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C
$$\displaystyle \frac {5h} {8} $$
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D
$$\displaystyle \frac {3h^2} {8R} $$
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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 
$$I=\frac { 1 }{ 6 } M{ L }^{ 2 }\\ \Rightarrow I=\frac { 1 }{ 6 } \times \frac { 2{ M } }{ \sqrt { 3 } \pi  } \times \left( \frac { 2R }{ \sqrt { 3 }  }  \right) ^{ 2 }=\frac { 4MR^{ 2 } }{ 9\sqrt { 3 } \pi  } $$ 
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#299440
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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
$$\displaystyle \frac {MR^2} {32 \sqrt{2} \pi} $$
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B
$$\displaystyle \frac {MR^2} {16 \sqrt{2} \pi} $$
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C
$$\displaystyle \frac {4MR^2} {9 \sqrt{3} \pi} $$
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D
$$\displaystyle \frac {4MR^2} {3 \sqrt{3} \pi} $$
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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) $$=-\dfrac{3GM_c}{2R_c}$$
$$V=-\dfrac{11GM}{8R}-(-\dfrac{3GM_c}{2R_c})=-\dfrac{11GM}{8R}+\dfrac{3G\dfrac{M}{8}}{2\dfrac{R}{2}}\\\Rightarrow V= -\dfrac{GM}{R}$$
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#299463
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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
$$\displaystyle \frac {-GM} {2R} $$
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B
$$\displaystyle \frac {-GM} {R} $$
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C
$$\displaystyle \frac {-2GM} {3R} $$
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D
$$\displaystyle \frac {-2GM} {R} $$
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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
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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
$$\displaystyle \left[ \left( \frac {T_M} {T} \right)^2 - 1 \right] \frac {A} {M_g} $$
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B
$$\displaystyle \left[ \left( \frac {T_M} {T} \right)^2 - 1 \right] \frac {M_g} {A} $$
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C
$$\displaystyle \left[ 1 - \left( \frac {T_M} {T} \right)^2 \right] \frac {A} {M_g} $$
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D
$$\displaystyle \left[ 1 - \left( \frac {T} {T_M} \right)^2 \right] \frac {A} {M_g} $$
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9
#299516
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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
$$ T \propto e^{-R} $$
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B
$$ T \propto e^{-3R} $$
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C
$$\displaystyle T \propto \dfrac {1} {R} $$
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D
$$\displaystyle T \propto \dfrac {1} {R^3} $$
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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 
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#299541
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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
$$ \ln2 , 4\ln2 $$
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B
$$ \ln2 , \ln2 $$
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C
$$ \ln2 , 2\ln2 $$
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D
$$ 2\ln2 , 8\ln2 $$
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