Basic Science Test 11

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Basic Science Test 11

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Question 1
1 / -0

In the arrangement shown in figure wedge of mass $$M$$ moves towards left with an acceleration $$a$$. All surfaces are smooth. The acceleration of mass $$m$$ relative to wedge is:

A
$$a/2$$

B
$$\cfrac{2Ma}{m}$$

C
$$\cfrac{2(M+m)a}{m}$$

D
$$\cfrac{(M+m)a}{m}$$

Solution

Force on $$(M+m)$$ in backward direction $$=(M+m)a$$
acceleration of $$m$$ in forward direction $$=\cfrac{(M+m)a}{m}$$
Question 2
1 / -0

Figure shows the graph of x-coordinate of a particle moving along x-axis as function of time. Average velocity during $$t=0$$ to $$6s$$ and instantaneous velocity at $$t=3s$$ respectively, will be

A
$$10m/s,0$$

B
$$60m/s,0$$

C
$$0,0$$

D
$$0,10m/s$$

Solution

Velocity= slope of x-t graph=0 (at $$t=35$$)
Net displacement from $$t=0$$ to $$t=6s=0$$
Ave. velocity$$=\cfrac{Net\quad displacement}{time}=\cfrac{0}{6}=0$$
Question 3
1 / -0

A bob of a simple pendulum of length $$2m$$ is moved sideways and whirled in horizontal circle of radius $$1.732m$$. If the mass of the bob is $$2kg$$, then tension in the string is

A
$$34.5N$$

B
$$22.8N$$

C
$$39.2N$$

D
$$57.6N$$

Solution

$$\sin\theta=\frac{1.732}{2}$$
$$\sin\theta=\frac{\sqrt{3}}{2}$$
$$\theta=60^0$$
$$T\cos\theta=mg$$
$$T=\frac{mg}{\cos\theta}=\frac{2*9.8}{\cos60^0}=\frac{19.6}{\frac{1}{2}}=39.8N$$
Question 4
1 / -0

A particle is in uniform circular motion, then its velocity is perpendicular to

A
Net force

B
Centripetal acceleration

C
Angular velocity

D
All of these

Solution

$$V\bot a_c$$
if in uniform circular motion then only centripetal force occurs
net force =centripetal force
so netforce $$\bot a_v$$
Question 5
1 / -0

A wheel of mass $$40kg$$ and radius of gyration $$0.5m$$ comes to rest from a speed of $$1800r\pm $$ in $$30s$$. Assuming that the retardation is uniform, the value of the retarding torque in $$Nm$$, is

A
$$10\pi$$

B
$$20\pi$$

C
$$30\pi$$

D
$$40\pi$$

Solution

$$\tau =I\alpha$$
$$=(m{k}^{2}).\alpha$$
$$=40\times {(0.5)}^{2}\times 2\pi$$
$$=20\pi$$ rad/s
Question 6
1 / -0

What is the minimum coefficient of friction for a solid sphere to roll without slipping on an inclined plane of inclination $$\theta$$?

A
$$\frac{2}{7}\tan\theta$$

B
$$\frac{1}{3}g\tan\theta$$

C
$$\frac{1}{2}\tan\theta$$

D
$$\frac{2}{5}\tan\theta$$

Solution

Here we are talking about minimum coefficient of friction this suggests that friction acting would be $$\mu mg\cos\theta$$
taking moment about centre,
$$\mu mg\cos\theta*r=I*\alpha$$
or, $$=\frac{\mu mg\cos\theta_{r}^2}{I}$$
now, from force balance,
$$mg\sin\theta-\mu mg\cos\theta=ma$$
$$a=g\sin\theta-\mu g\cos\theta$$
from condition of rolling we know that
$$a=r\alpha$$
$$\frac{\mu mg\cos\theta_{r}^2}{I}=g\sin\theta-\mu mg\cos\theta$$
simplifying this, we get
$$\mu=\frac{\tan\theta}{1+\frac{mr^2}{I}}$$
$$I=\frac{2}{5}mr^2$$
Putting the value of moment of ionertia in expression of coefficient of friction we get,
$$\mu=\frac{2}{7}\tan\theta$$
Question 7
1 / -0

A body is started from rest with acceleration $$2 m/s^{2}$$ till it attains the maximum velocity then retards to rest with 3 $$m/s^{2}$$ It total time taken is 10 second then maximum speed attained is

A
$$12 m/s$$

B
$$8 m/s$$

C
$$6 m/s$$

D
$$4 m/s$$
Question 8
1 / -0

A partice is at rest and it is at origin $$(0,0)$$ at $$t=0$$. If velocity varies as $$v=k{x}^{2}$$ where $$v$$ is the velocity and $$x$$ is displacement. Find how much distance it will travel in $$2$$ sec

A
$$5m$$

B
$$2m$$

C
$$0$$

D
none of these

Solution

$$V=k{x}^{2}$$
$$\cfrac{dx}{dt}=k{x}^{2}$$
$$dx=k{x}^{2}dt$$
$$\int _{ 0 }^{ x }{ \cfrac { 1 }{ { x }^{ 2 } } } =\int _{ 0 }^{ 2 }{ x } dt$$
$$-\cfrac{1}{x}={ \left[ kt \right] }_{ 0 }^{ 2 }$$
$$-\cfrac{1}{x}=k2$$
$$x=-\cfrac{1}{2k}$$
$$-\cfrac{1}{x}=kt+k$$
at $$t=0$$ and $$d=0$$
$$k=\infty$$
$$x=-\cfrac{1}{2\infty}$$
$$k=0$$
Question 9
1 / -0

Two identical rings each of mass m with their planes mutually perpendicular , radius R are welded at their point of contact O . If the system is free to rotate about an axis passing through the point p perpendicular to the plane of the paper the moment of inertia of the system about this axis equal to :

A
$$ 6.5 m R^2$$

B
$$12m R^2$$

C

$$6 m R^2$$

D
$$11.5 m R^2$$

Solution

Answer
$$ I = I_{1} + I_{1} $$
$$ I_{1} = \dfrac{mr^{2}}{2} + M (2R + R)^{2} = 9.5MR^{2}$$
$$ I_{2} = \dfrac{mr^{2}}{2} + MR^{2} = 2 MR^{2} $$
$$ I = I_{1} + I_{2} $$
$$ I = 9.5Mr^{2} + M R^{2} $$
$$ I= 11.5MR^{2}$$
Question 10
1 / -0

A body is thrown up with a velocity $$29.23ms!$$ $$V$$ distance travelled in last second of upward motion is

A
$$2.3m$$

B
$$6m$$

C
$$9.8m$$

D
$$4.9m$$

Solution

Total time of motion:
$$v=u+at$$
$$x_0=29.23-9.8\times 6$$
$$xt=2.98\ sec$$
distance total
$$\Rightarrow v^2=u^2+2as$$
$$\Rightarrow 0=(29.23)^2=2\times 9.8 \times s$$
$$\Rightarrow s=43.59\ m$$
Now in $$1.96\ sec$$
$$s=29.3\times 1.98-\dfrac{1}{2}\times 9.8 (1.98)^2$$
$$s=36.6654$$
in last $$s= 43.59-38.66$$
$$=4.93m$$