Basic Science Test 27

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Basic Science Test 27

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Question 1
1 / -0

The direction of projection of particle is shown in the figure for an observer on trolley. An observer on the ground sees the ball rise vertically. The maximum height reached by ball as seen from the trolley is

A
$$10m$$

B
$$15m,$$

C
$$20m$$

D
$$5m$$

Solution

Horizontal V = 0
Vcos60 - 10=0

V x 1/2 =10

V = 20m/s

Vertical Velocity = Vsin60

20 x ((3)^-1/2)/2) = 10((3)^1/2) m/s

2ay = v^2 -u^2

2 x (-10) x h = 0 - 10((3)^1/2)^2

-20h = -300

h= 15m

OR

$$10-v\cos60^0=0$$
$$v=20m/s$$
$$H=\frac{v^2\sin^260^0}{2g}=15m$$
Question 2
1 / -0

In the graph shown the displacement of the body is $$5$$ seconds will be

A
$$60\ m $$

B
$$40\ m $$

C
$$80\ m $$

D
$$100\ m $$

Solution

Displacement = $$ \left (\dfrac{dx}{dt} \right ) \times dt = \int dx $$
= Area under V-t curve
$$ = A_{1} + A_{2} + A_{3} + A_{4} + A_{5} $$
$$ = \dfrac{1}{2} [40 - 20] \times (2 - 0) + (3 - 2) \times 20 + \dfrac{1}{2} \times 20 \times (4 - 3) + \dfrac{1}{2} \times (-20) \times (5 - 4) + 2 \times 20 $$

Disp = $$20 + 20 + 10 - 10 + 40 = 40\ m + 40\ m $$
$$ = 80\ m $$
Question 3
1 / -0

A satellite of mass $$m$$ is orbiting the earth (or radius $$R$$) at a height $$h$$ from its surface. The total energy of the satellite in terms of $$g_0$$ the value of acceleration due to gravity at the earth's surface is

A
$$\frac{mg_0R^2}{2(R+h)}$$

B
$$-\frac{mg_0R^2}{2(R+h)}$$

C
$$\frac{2mg_0R^2}{R+h}$$

D
$$-\frac{2mg_0R^2}{R+h}$$

Solution

$$r=R+h$$
$$P.E=-\frac{GMm}{r}$$
$$v=\sqrt{\frac{Gm}{r}}$$
$$K.E=\frac{1}{2}m*\frac{GM}{r}=\frac{GMm}{2r}$$
$$T.E=P.E+K.E$$
$$=-\frac{GMm}{2r}$$
$$=-\frac{Gmn}{2(R+h)}$$
$$=-\frac{Gm}{R^2}*\frac{R^2m}{2(R+h)}=-\frac{G_0mR^2}{2(R+h)}$$
Question 4
1 / -0

A car moving along a straight highway with speed $$144$$ km/h is brought to stop within a distance of $$400\ m $$ with uniform retardation. How long does it take for the car to stop ?

A
$$10\ s $$

B
$$15\ s $$

C
$$20\ s $$

D
$$25\ s $$

Solution

$$v^{2} = u^{2} + 2as $$
$$a = \dfrac{-u^2}{2s} = -(144 \times \dfrac{5}{18}) = -2\ m/sec^{2} $$

$$v = u + at $$
$$t = \dfrac{-u}{a} = -\dfrac{40}{-2} = 20 sec $$
Question 5
1 / -0

Refer to the system shown above. the ratio of tension $$T_1$$ and $$T_2$$ is

A
$$\frac{m_1}{m_1+m_2}$$

B
$$\frac{m_2}{m_1+m_2}$$

C
$$\frac{m_1}{m_2}$$

D
$$\frac{m_2}{m_1}$$

Solution

$$a=(\frac{T_2}{m_1+m_2})$$

$$T_1=m_1a$$

$$T_1=\frac{m_1\cdot T_2}{m_1+m_2}$$

$$\frac{T_1}{T_2}=(\frac{m_1}{m_1+m_2})$$
Question 6
1 / -0

Determine force on P in the following system of the charges q kept vertices of an equilateral triangle of side 'a'.

A
$$ \sqrt{2} \dfrac{kq^2}{a^2} $$

B
$$ \sqrt{3} \dfrac{kq^2}{a^2} $$

C
$$ \dfrac{kq^2}{a^2} $$

D
$$ \dfrac{4kq^2}{a^2} $$

Solution

$$F_{net} = \sqrt{F^{2} + F^{2} + 2 F^{2} \cos 60} $$

$$ = \sqrt{F^{2} + F^{2} + F^{2}} $$

$$ = \sqrt{3} F $$

$$ = \sqrt{3} \dfrac{kq^{2}}{a^2} $$
Question 7
1 / -0

The ratio of the weight of a man in a stationary lift and when it is moving downward with uniform acceleration 'a' is $$3:2$$. The value of 'a' is (g = acceleration due to gravity)

A
$$(3/2)g$$

B
$$g$$

C
$$(2/3)g$$

D
$$g/3$$

Solution

The weight of man when he is stationary is $$mg$$
The weight of man when he is moving downward with acceleration $$a$$ is $$m(g-a)$$
The ratio is $$\frac{g}{g-a}=\frac{3}{2}$$
$$2g=3g-3a$$
$$a=\frac{g}{3}$$
Therefore the correct option is (D).
Question 8
1 / -0

$$1$$ amu is

A
$$1.66 \times 10^{-24}kg$$

B
$$1.66 \times 10^{-27}kg$$

C
$$\dfrac{1}{NA}$$

D
Both (B) and (C)

Solution

$$1$$ amu = $$1.66 \times 10^{-24}gm=1.66\times 10^{-27}kg$$
Weight of $$1$$ element or atom
$$1$$ mole = $$N_A$$ atom = Atomic mass
So, weight of one atom = $$(\dfrac{1}{N_A})$$ called amu.
Question 9
1 / -0

Two light strings of length $$4cm$$ and $$3cm$$ are tied to a bob of weight $$500gm$$. The free ends of the strings are tied to pegs in the same horizontal line and separated by $$5cm$$. The ratio of tension in the longer string to that in the shorter string is

A
$$4:3$$

B
$$3:4$$

C
$$4:5$$

D
$$5:4$$

Solution

Let the two pegs be 'B' and 'C' and the bob be at 'A'.
Let the tension in the 2 strings be $$T_1$$ and $$T_2$$
$$T_1\cos\theta=T_2\sin\theta$$
$$\frac{T_1}{T_2}=\cot\theta=\frac{3}{4}$$
Question 10
1 / -0

A bob is suspended from a ideal string of length 'l'. Now it is pulled to a side through $$60^0$$ to vertical and whirled along a horizontal circle. Then its period of revolution is

A
$$\pi\sqrt{l/g}$$

B
$$\pi\sqrt{l/2g}$$

C
$$\pi\sqrt{2l/g}$$

D
$$2\pi\sqrt{l/g}$$

Solution

From F.D
$$T=60^0=mg$$
$$T*\frac{1}{2}=mg$$
$$T=2mg$$....(i)
$$\sin60^0=mdw^2$$
$$mg*\frac{\sqrt{3}}{2}=m*l\sin60^0*\frac{4\pi^2}{T^2}$$
$$T^2=l\frac{4\pi^2}{2g}$$
$$T=\pi\sqrt{\frac{2l}{g}}$$