Physics Test

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Physics Test - 47

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Question 1
1 / -0

Let 'M' be the mass and 'L' be the length of a thin uniform rod. In first case, axis of rotation is passing through centre and perpendicular to the length of the rod. In second case axis of rotation is passing through one end and perpendicular to the length of the rod. The ratio of radius of gyration in first case to second case is :

A
1

B
1/2

C
1/4

D
1/8

Solution

Moment of inertia of the rod about an axis passing through its centre and perpendicular to the length \(I_{1}=\frac{1}{12} M L^{2}\) Using \(M k_{1}^{2}=I_{1}=\frac{1}{12} M L^{2}\)
We get radius of gyration \(k_{1}=\frac{L}{\sqrt{12}}\)

Moment of inertia of the rod about an axis passing through its one end and perpendicular to the length \(I_{2}=\frac{1}{3} M L^{2}\)

Using \(M k_{2}^{2}=I_{2}=\frac{1}{3} M L^{2}\)
We get radius of gyration \(k_{2}=\frac{L}{\sqrt{3}}\)

Thus ratio of radius of gyration \(\frac{k_{1}}{k_{2}}=\frac{L / \sqrt{12}}{L / \sqrt{3}}\) \(\Longrightarrow \frac{k_{1}}{k_{2}}=\frac{1}{2}\)
Hence, the correct option is (B)
Question 2
1 / -0

Four small objects each of mass m are fixed at the corners of a rectangular wire-frame of negligible mass and of sides a and b (a>b). If the wire frame is now rotated about an axis passing along the side of length b, then the moment of inertia of the system for this axis of rotation is:

A
2ma²

B
4ma²

C
2m (a²+b²)

D
2m (a²-b²)

Solution

The given situation can be shown as given

Moment of inertia of the system about side of length b say CD is

= M.I of mass at A about CD+ M.I of mass at B about CD+ M.I of mass at C about CD+ M.I of mass at D about CD

=m(a)²+m(a)²+m(0)²+m(0)²

=2ma²
Hence, the correct option is (A)
Question 3
1 / -0

A metal bar of 70 cm long and 4 kg in mass supported on two knife edges placed 20 cm from each end. A 6 kg load is suspended at 30 cm from on end. Find the normal reaction at the knife-edge. (assume it to be of uniform cross section and homogeneous).

A
\(55 \mathrm{N}\) at \(K_{1}\) and \(33 \mathrm{N}\) at \(K_{2}\)

B
\(45 \mathrm{N}\) at \(K_{1}\) and \(53 \mathrm{N}\) at \(K_{2}\)

C
\(55 \mathrm{N}\) at \(K_{1}\) and \(43 \mathrm{N}\) at \(K_{2}\)

D
\(43 \mathrm{Nat} K_{1}\) and \(55 \mathrm{N}\) at \(K_{2}\)

Solution

Let the normal reactions from the knife edges \(\left(K_{1} \& K_{2}\right)\) be \(N_{1}, N_{2}\)
Now since the bar is in equilibrium, \(N_{1}+N_{2}=m g\)
\(N_{1}+N_{2}=10 \times 9.8 N=98 N\)
Also the system is in rotational equilibrium too. Thus, about the center of the rod, the net torque must be zero. Hence \((6 \times 9.8)(5)+\left(N_{2}\right)(25)=\left(N_{1}\right)(25)\)
\(\Longrightarrow N_{1}=55 N\)
\(N_{2}=43 N\)
Hence normal reaction \(55 \mathrm{N}\) at \(K_{1}\) and \(43 \mathrm{N}\) at \(K_{2}\) are formed.
Hence, the correct option is (C)
Question 4
1 / -0

A round uniform body of radius R, mass M and moment of inertia I, rolls down (without slipping) an inclined plane making an angle with the horizontal. Then its acceleration is :

A
\(\frac{g \sin \theta}{1+I / M R^{2}}\)

B
\(\frac{g \sin \theta}{1+M R^{2} / I}\)

C
\(\frac{g \sin \theta}{1-I / M R^{2}}\)

D
\(\frac{g \sin \theta}{1-M R^{2} / I}\)

Solution

For translational motion: \(F=M a\)
\(m g \sin \theta-f=m a\)
(1) For rotational motion: Net torque about \(\mathrm{O}, \tau=I \alpha\) \(f R=I \alpha\)
\(\Longrightarrow f=\frac{I \alpha}{R}\)
Also body rolls without slipping, thus \(a=r \alpha\) Hence, \(f=\frac{I a}{R^{2}}\)
Purting it in ( 1\(), m g \sin \theta-\frac{I a}{R^{2}}=m a\)
\(\Rightarrow a=\frac{g \sin \theta}{1+\frac{I}{M R^{2}}}\)
Hence, the correct option is (A)
Question 5
1 / -0

A uniform cube of side 'b' and mass M rest on a rough horizontal table. A horizontal force F is applied normal to one of the face at a point, at a height 3b/4 above the base. What should be the coefficient of friction (μ) between cube and table so that it will tip about an edge before it starts slipping?

A
\(\mu>\frac{2}{3}\)

B
\(\mu>\frac{1}{3}\)

C
\(\mu>\frac{3}{2}\)

D
none

Solution

For equilibrium in y direction we have \(N=M g\)
For equilibrium in x direction we have \(F=f\) (here \(f\) is the frictional force) or \(F=\mu N=\mu M g\)
For equilibrium of torque about center we have \(N \times x=f \times \frac{b}{2}+F \times \frac{b}{4}\)
or \(M g \times \frac{b}{2}=\mu M g \times \frac{b}{2}+\mu M g \times \frac{b}{4}\left(\right.\) as \(x=\frac{b}{2}\) for toppling \()\)
or \(\mu=\frac{2}{3}\)
Thus for tipping we have \(\mu>\frac{2}{3}\)
Hence, the correct option is (A)
Question 6
1 / -0

Two persons P and Q of the same height are carrying a uniform beam of length 3m. If Q is at one end, then the distance of P from the other end such that P, Q receive loads in the ratio 5:3 is:

A
0.5 m

B
0.6 m

C
0.75 m

D
1 m

Solution

\(N_{P}+N_{Q}=M g\)
also \(\frac{N_{P}}{N_{Q}}=\frac{5}{3}\)
\(\Rightarrow N_{P}=\frac{5}{8} M g, N_{Q}=\frac{3}{8} M g\)
Calculating torque about the end opposite to \(Q\) \(N_{P} d_{1}+N_{Q}\left(3-d_{2}\right)=M g \frac{3}{2}\)
\(\frac{5}{8} M g d_{1}+\frac{3}{8} M g \times 3=M g \frac{3}{2}\)
\(\therefore \frac{5}{8} M g d_{1}+\frac{3}{8} M g \times 3=M g \frac{3}{2}\)
\(\frac{5}{8} M g d_{1}=\left(\frac{12-9}{8}\right) M g\)
\(d_{1}=\frac{3}{5}=0.6 m\)
Hence, the correct option is (B)
Question 7
1 / -0

A solid cylinder is projected on a rough surface having coefficient of friction μ, with velocity v₀and angular velocity ω₀=v₀/2r, (whre r is radius). What is the time t at which rolling without slipping occurs is :

A
\(\frac{v_{0}}{3 \mu g}\)

B
\(\frac{v_{0}}{4 \mu g}\)

C
\(\frac{v_{0}}{2 \mu g}\)

D
\(\frac{v_{0}}{\mu g}\)

Solution

Rolling without slipping will occur at \(v_{o}=r \omega\) or at \(\omega=\frac{v_{o}}{r}\) Also using torque equilibrium we have \(\mu m g r=I \alpha\) or \(\mu m g r=\frac{1}{2} m r^{2} \alpha\)
or \(\alpha=\frac{2 \mu g}{r}\)
Now using the equation \(\omega_{f}=\omega_{i}+\alpha t\) we have \(\frac{v_{o}}{r}=\frac{v_{o}}{2 r}+\frac{2 \mu g}{r} t\)
Or
\(t=\frac{v_{o}}{4 \mu g}\)
Hence, the correct option is (B)
Question 8
1 / -0

If the Earth shrinks such that its density becomes 8 times its present value then the new duration of the day in hours will be : (assume that there is no change in axis of rotation of the earth)

A
24

B
12

C
6

D
3

Solution

If \(M\) is mass of the earth, then clearly \(M_{i}=M_{f}\) (where \(M_{i}\) and \(M_{f}\) represent the initial and final mass of the earth) Now \(d=\frac{M}{V},\) where \(\mathrm{V}\) will be the volume of a sphere, given by \(V=\frac{4}{3} \pi R^{3},\) where \(\mathrm{R}\) is the radius of earth. since Mass is constant, \(d_{i} R_{i}^{3}=d_{f} R_{f}^{3}\) putting \(d_{f}=8 d_{i},\) we get: \(R_{f}=\frac{R_{i}}{2}\) Now since there is no external torque applied, by conservation of angular momentum, we get:
\(L_{i}=L_{f}\)
hence, \(I_{i} \omega_{i}=I_{f} \omega_{f}\)
hence, \(\frac{2}{5} M_{i} R_{i}^{2} \frac{2 \pi}{T_{i}}=\frac{2}{5} M_{f} R_{f}^{2} \frac{2 \pi}{T_{f}}\)
putting \(M_{i}=M_{f}, T_{i}=24 h, R_{f}=\frac{R_{i}}{2}\)
we get \(T_{f}=6 h\)
Hence, the correct option is (C)
Question 9
1 / -0

Keeping the mass of earth as constant, if its radius is reduced to 1/4th of its initial value, then the period of revolution of earth about its own axis, and passing through the centre, in hrs, is:

(Assume earth to be a solid sphere and its initial period of rotation as 24 hrs)

A
12

B
3

C
6

D
1.5

Solution

Assuming earth to be a solid sphere, its moment of inertia \(I=\frac{2}{5} M R^{2}\)
As no external torque is acting on earth, thus its angular momentum is constant, i.e, \(I w=\)constant \(\therefore \frac{2}{5} M R^{2} \times \frac{2 \pi}{T}=\) constant \(\quad\) where \(T\) is the time period
of earth's rotation
\(\Longrightarrow \frac{R_{1}^{2}}{T_{1}}=\frac{R_{2}^{2}}{T_{2}} \quad(\because M=\)constant \()\)
Given : \(\quad R_{1}=R\)
\(R_{2}=\frac{R}{4} \quad T_{1}=24 h r s\)
\(\therefore \quad \frac{R^{2}}{24}=\frac{\frac{R^{2}}{16}}{T_{2}} \quad \Longrightarrow T_{2}=\frac{24}{16}=1.5 \mathrm{hrs}\)
Hence, the correct option is (D)
Question 10
1 / -0

A uniform meter scale of mass 1kg is placed on table such that a part of the scale is beyond the edge. If a body of mass 0.25kg is hung at the end of the scale then the minimum length of scale that should lie on the table so that it does not tilt is :

A
90cm

B
80cm

C
70cm

D
60cm

Solution

the normal will shift to the edge of table to prevent toppling.

Torque about edge,

mg(1−d−0.5m)=(0⋅25)g d

4(0.5−d)=d

2−4d=d

2/5=d=0⋅4m

part lying on table: =1−⋅4=0.6m

Hence, the correct option is (D)

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Physics Test - 47

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Physics Test - 47

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SHARING IS CARING

Question 1

1

1/2

1/4

1/8

Solution

Question 2

2ma2

4ma2

2m (a2+b2)

2m (a2-b2)

Solution

Question 3

\(55 \mathrm{N}\) at \(K_{1}\) and \(33 \mathrm{N}\) at \(K_{2}\)

\(45 \mathrm{N}\) at \(K_{1}\) and \(53 \mathrm{N}\) at \(K_{2}\)

\(55 \mathrm{N}\) at \(K_{1}\) and \(43 \mathrm{N}\) at \(K_{2}\)

\(43 \mathrm{Nat} K_{1}\) and \(55 \mathrm{N}\) at \(K_{2}\)

Solution

Question 4

\(\frac{g \sin \theta}{1+I / M R^{2}}\)

\(\frac{g \sin \theta}{1+M R^{2} / I}\)

\(\frac{g \sin \theta}{1-I / M R^{2}}\)

\(\frac{g \sin \theta}{1-M R^{2} / I}\)

Solution

Question 5

\(\mu>\frac{2}{3}\)

\(\mu>\frac{1}{3}\)

\(\mu>\frac{3}{2}\)

none

Solution

Question 6

0.5 m

0.6 m

0.75 m

1 m

Solution

Question 7

\(\frac{v_{0}}{3 \mu g}\)

\(\frac{v_{0}}{4 \mu g}\)

\(\frac{v_{0}}{2 \mu g}\)

\(\frac{v_{0}}{\mu g}\)

Solution

Question 8

24

12

6

3

Solution

Question 9

12

3

6

1.5

Solution

Question 10

90cm

80cm

70cm

60cm

Solution

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2ma²

4ma²

2m (a²+b²)

2m (a²-b²)