Physics Test

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Physics Test - 48

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Question 1
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Two masses m₁ and m₂ are connected by a string of length l. They are held in a horizontal plane at a height H above two heavy plates A and B made of different material placed on the floor. Initially distance between two masses is a < l. When the masses are released under gravity they make collision with A and B co - efficient of restitution 0.8 and 0.4 respectively. The time after the collision when the string becomes tight is :

A
$\frac{5}{2} \sqrt{\frac{l^{2}-a^{2}}{2 g H}}$

B
$\sqrt{\frac{2 \mathrm{g}}{\mathrm{H}}}$

C
$\frac{3}{2} \sqrt{\frac{l^{2}-a^{2}}{2 g H}}$

D
None of these

Solution

String will become tight when
$\mathrm{h}_{2}-\mathrm{h}_{1}=\sqrt{l^{2}-\mathrm{a}^{2}}$
$\mathrm{h}_{1}=0.4 \mathrm{vt}-\frac{1}{2} \mathrm{gt}^{2} \& \mathrm{h}_{2}=0.8 \mathrm{vt}-\frac{1}{2} \mathrm{gt}^{2}$
$\mathrm{h}_{2}-\mathrm{h}_{1}=0.4 \mathrm{vt}$
$\therefore 0.4 \mathrm{vt}=\sqrt{l^{2}-\mathrm{a}^{2}},$ Where $\mathrm{v}=\sqrt{2 \mathrm{gH}}$

$t=\frac{5}{2} \sqrt{\frac{l^{2}-a^{2}}{2 g H}}$
Concepts :
Main Concept :
Elastic and Inelastic CollisionsA perfectly elastic collision is defined as one in which there is no loss of kinetic energy in the collision. An inelastic collision is one in which part of the kinetic energy is changed to some other form of energy in the collision. Any macroscopic collision between objects will convert some of the kinetic energy into internal energy and other forms of energy, so no large scale impacts are perfectly elastic. Momentum is conserved in inelastic collisions, but one cannot track the kinetic energy through the collision since some of it is converted to other forms of energy. Collisions in ideal gases approach perfectly elastic collisions, as do scattering interactions of sub-atomic particles which are deflected by the electromagnetic force. Some large-scale interactions like the slingshot type gravitational interactions between satellites and planets are perfectly elastic.

Collisions between hard spheres may be nearly elastic, so it is useful to calculate the limiting case of an elastic collision. The assumption of conservation of momentum as well as the conservation of kinetic energy makes possible the calculation of the final velocities in two-body collisions.

An elastic collision is defined as one in which both conservation of momentum and conservation of kinetic energy are observed. This implies that there is no dissipative force acting during the collision and that all of the kinetic energy of the objects before the collision is still in the form of kinetic energy afterward.
For macroscopic objects which come into contact in a collision, there is always some dissipation and they are never perfectly elastic. Collisions between hard steel balls as in the swinging balls apparatus are nearly elastic.
"Collisions" in which the objects do not touch each other, such as Rutherford scattering or the slingshot orbit of a satellite off a planet, are elastic collisions. In atomic or nuclear scattering, the collisions are typically elastic because the repulsive Coulomb force keeps the particles out of contact with each other.
Collisions in ideal gases are very nearly elastic, and this fact is used in the development of the expressions for gas pressure in a container.
Coefficient of restitution
The coefficient of restitution (COR) is a measure of the "bounciness" of a collision between two objects: how much of the kinetic energy remains for the objects to rebound from one another vs. how much is lost as heat, or work done deforming the objects.
The coefficient, e is defined as the ratio of relative speeds after and before an impact, taken along the line of the impact:
Coefficient of restitution(e) = Relative speed after collision/Relative speed before collision
Hence, the correct option is (A)
Question 2
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A block of weight $5 N$ is pushed against a vertical wall by a force $12 N$. The coefficient of friction between the wall and block is $0.6 .$ Find the magnitude of the force exerted by the wall on the block:

A
13 N

B
12N

C
16N

D
0 N

Solution

Reaction $=\sqrt{(12)^{2}+5^{2}}=\sqrt{169}$
Hence, the correct option is (A)
Question 3
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Calculate the gas constant for $1 g$ of a gas from the following data: $c_{p}=0.245 \mathrm{cal} g^{-1^{\circ}} \mathrm{C}^{-1} ; \mathrm{c}_{v}=0.165 \mathrm{cal} \mathrm{g}^{-10} \mathrm{C}^{-1}$ and $\mathrm{j}=4.2 \times$ 107 erg cal ^-1

A
3.36 × 10⁶ erg g^-1 ^oC^-1

B
2.13 × 10⁶ erg g^-1 ^oC^-1

C
4.26 × 10⁶ erg g^-1 ^oC^-1

D
2.57 × 10⁶ erg g^-1 ^oC^-1

Solution

Now, $c_{p}-c_{v}=\frac{r}{J} \quad$ or $\quad r=J\left(c_{p}-c_{v}\right)$
$$
\begin{aligned}
& r=4.2 \times 10^{7}(0.245-0.165) \\
&=3.36 \times 10^{6} \mathrm{erg} \mathrm{g}^{-1}{ }^{\circ} \mathrm{C}^{-1}
\end{aligned}
$$Concepts:
Main Concept :
Relation Between Molar Specific Heat at Constant Pressure and Volume Specific heat of gas during isobaric process: $\mathrm{C}_{\mathrm{p}}=\left(\frac{\mathrm{f}}{2}+1\right) \mathrm{R}$
Specific heat of gas during isochoric process : $\mathrm{C}_{\mathrm{V}}=\left(\frac{\mathrm{f}}{2}\right) \mathrm{R}$
This gives $\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}=\mathrm{R}$
Hence, the correct option is (A)
Question 4
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A copper sphere of 10 cm diameter is lowered into a water filled hemispherical copper vessel of 20 cm diameter so that the sphere and the vessel becomes concentric. Electrical conductivity of water isσ = 10^-3S/m. The electrical resistance between the sphere and the vessel is:

A
1450 Ω

B
1682.4 Ω

C
1489.6 Ω

D
1591.6 Ω

Solution

The arrangerment is shown in figure. consider the hemispherical shell of radius 'rive thickness dras Shown. Resistance of this strell is : $d R=\frac{d q}{\sigma \times 2 \pi q^{2}}$ Integrating both sides

$\begin{aligned} R=& \frac{1}{2 \pi \sigma} \int_{x=5}^{x=10(m) d x} \\ &=1591.6 \Omega \end{aligned}$
Hence, the correct option is (D)
Question 5
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If a gas is heated at constant pressure then what percentage of total heat supplied is used for doing work ? (Given : for gas = 4/3)

A
25%

B
50%

C
75%

D
80%

Solution

Heat supplied $\mathrm{dQ}=\mathrm{nC}_{\mathrm{p}} \mathrm{d} \mathrm{T}$
Heat used for work $=\mathrm{dW}=$ nRdT
$\frac{\mathrm{dW}}{\mathrm{dQ}}=\frac{\mathrm{R}}{\mathrm{C}_{\mathrm{P}}}=\frac{\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}}{\mathrm{C}_{\mathrm{P}}}=1-\frac{\mathrm{C}_{\mathrm{V}}}{\mathrm{C}_{\mathrm{P}}}$
$=\left(1-\frac{1}{\gamma}\right)=\left(1-\frac{3}{4}\right)=\frac{1}{4}$
$=25 \%$
Concepts:
Main Concept:
Thermodynamics Process
The process of change of state of a system involves change of thermodynamic variables such as pressure $P$, volume $\mathrm{V}$ and temperature T of the system. This process is known as thermodynamic process.
Hence, the correct option is (A)
Question 6
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A wet open umbrella is held vertical and it whirld about the handle at a uniform rate of 21 revolutions in 44s. If the rim of the umbrella is circle of 1m in diameter and the height of the rim above the flour is 4.9m, the locus of the drop is a circle of radius.

A
√2.5m

B
1m

C
3m

D
1.5m

Solution

From equation of motion $h=u t+\frac{1}{2} g t^{2}$
where u is initial velocity and t is time since, $u=0$ $\therefore t=\sqrt{\frac{2 h}{g}}=\frac{\sqrt{2 \times 4.9}}{9.8}=1 s$
The horizontal range of the drop $=x$, then $x=\left(\frac{v_{t}}{o}\right) t$
Also, $\omega=\frac{\Delta \theta}{\Delta t}=\frac{21 \times 2 \pi}{44}=3 \mathrm{rad} / \mathrm{s}$
Tangential speed $v_{t}=r \omega=0.5 \times 3 \times 1.5 \mathrm{m} / \mathrm{s}$
$x=1.5 \times 1=1.5 m$
Locus of drop $=\sqrt{x^{2}+r^{2}}=\sqrt{(1.5)^{2}+(0.5)^{2}}$
$=\sqrt{2.5 \mathrm{m}}$
Hence, the correct option is (A)
Question 7
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Fig. shows a uniform solid block of mass M and edge lengths a, b and c . Its M.I about an axis passing through one corner and perpendicular (as shown) to the large face of the block is :

A
$\frac{M}{3}\left(a^{2}+b^{2}\right)$

B
$\frac{M}{4}\left(a^{2}+b^{2}\right)$

C
$\frac{7 M}{12}\left(a^{2}+b^{2}\right)$

D
$\frac{M}{12}\left(a^{2}+b^{2}\right)$

Solution

Moment of Inertia of plate passing through $O$ is

$I_{x}=M a^{2} / 12$ and, $I_{y}=M b^{2} / 12$
by perpendicular axis theorm $I_{z}=M a^{2} / 12+M b^{2} / 12$

The distance of edge and the center is $d=\sqrt{\left(a^{2}+b^{2}\right) / 4}$

So by parallel axis theorm, $I_{A}=M a^{2} / 12+M b^{2} / 12+M d^{2}=M\left(a^{2}+b^{2}\right) / 3$
Hence, the correct option is (A)
Question 8
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The radius of gyration of a body about an axis at a distance 6 cm from its centre of mass is 10 cm. Then, its radius of gyration about a parallel axis through its centre of mass will be :

A
80 cm

B
8 cm

C
0.8 cm

D
80 cm

Solution

Let radius of gyration for the axis not passing through center of mass be r and that for the axis passing through the center of mass be k and the distance between the two parallel axes be a.

Parallel axes theorem gives:

mr²=m(k²+a²⁾⇒r²=k²+a².

⇒k=√10²−6²=8cm.

Hence, the correct option is (B)
Question 9
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A uniform triangular plate ABC of moment of mass m and inertia I (about an axis passing through A and perpendicular to plane of the plate) can rotate freely in the vertical plane about point 'A' as shown in figure. The plate is released from the position shown in the figure. Line AB is horizontal. The acceleration of center of mass just after the release of plate is :

A
$\frac{m g a^{2}}{\sqrt{3} I}$

B
$\frac{m g a^{2}}{4 I}$

C
$\frac{m g a^{2}}{2 \sqrt{3} I}$

D
$\frac{m g a^{2}}{3 I}$

Solution

Torque about: $m g \frac{a}{2}=I \alpha$
$\Rightarrow \alpha=\frac{m g a}{2 I}$
$\Rightarrow A$ coleration $=\frac{a}{\sqrt{3}} \alpha=\frac{m g a^{2}}{2 \sqrt{3} I}$
Hence, the correct option is (C)
Question 10
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Identify the correct order in which the ratio of radius of gyration to radius increases for the following bodies.

I) rolling solid sphere

II) rolling solid cylinder

III) rolling hollow cylinder

IV) rolling hollow sphere

A
I,II,IV,III

B
I,III,II,IV

C
II,I,IV,III

D
II,I,III,IV

Solution

For solid sphere, the radius of gyration is $k=\sqrt{2 R^{2} / 5}=\sqrt{2 / 5} R$

For solid cylinder, the radius of gyration is $k=\sqrt{R^{2} / 2}=\sqrt{1 / 2} R$

For hollow cylinder, the radius of gyration is $k=\sqrt{R^{2}}=R$

For hollow sphere, the radius of gyration is $k=\sqrt{2 R^{2} / 3}=\sqrt{2 / 3} R$
Arranging in increasing order, \(IHence, the correct option is (A)