1

A bullet fired at a wooden block of thickness 0.15m manages to penetrate the block. If the mass of the bullet is 0.025kg and the average resisting force of the wood is 7.5 x 10^{3}N, calculate the speed of the bullet just before it hits the wooden block.

A

450ms^{-1}

B

400ms^{-1}

C

300ms^{-1}

D

250ms^{-1}

CORRECT OPTION:
c

F = ma

Therefore, a = \(\frac{F}{m}\)

a = \(\frac{7.5 \times 10^{3}}{0.025}\)

a = 3 x 10^{5}m/s^{2}

using V^{2} = U^{2} + 2ax;

V^{2} = 0^{2} + (2 x 3 x 10^{5} x 0.15)

Therefore V = 300 m/s

Therefore, a = \(\frac{F}{m}\)

a = \(\frac{7.5 \times 10^{3}}{0.025}\)

a = 3 x 10

using V

V

Therefore V = 300 m/s

2

The velocity of a sound wave at 27^{o}C is 360ms^{-1}. Its velocity at 127^{o}C is

A

120√3 ms^{-1}

B

240 ms^{-1}

C

240√3 ms^{-1}

D

720√3 ms^{-1}

CORRECT OPTION:
c

Velocity \(\alpha \sqrt{temperature}\)

\(\frac{V_1}{V_2} = \sqrt{\frac{T_1}{T_2}}\)

therefore, \(\frac{360}{V_2} = \sqrt{\frac{27 + 273}{127 + 273}}\)

\(V_2 = 240\sqrt{3}\)m/s

\(\frac{V_1}{V_2} = \sqrt{\frac{T_1}{T_2}}\)

therefore, \(\frac{360}{V_2} = \sqrt{\frac{27 + 273}{127 + 273}}\)

\(V_2 = 240\sqrt{3}\)m/s

3

The physical quantity that has the same dimensions as impulse is

A

energy

B

momentum

C

surface tension

D

pressure

CORRECT OPTION:
b

Impulse = F x t = m(v-u)

4

A ball is moving at 18ms^{-1} in a direction inclined at 60^{o} to the horizontal. The horizontal component to its velocity is

A

9√3ms^{-1} ·

B

6√3ms^{-1} ·

C

8√3ms^{-1} ·

D

9ms^{-1} ·

CORRECT OPTION:
d

5

In free fall, a body of mass 1kg drops from a height of 125m from rest in 5s. How long will it take another body of mass 2kg to fall from rest from the same height? [g = 10ms^{-1}·]

A

5s

B

10s

C

12s

D

15s

CORRECT OPTION:
a

In free fall, every object will reach the ground at same time

6

A ball of mass 0.15kg is kicked against a rigid vertical wall with a horizontal velocity of 50ms^{-1}. If it rebounced with a horizontal velocity of 30ms^{-1}, Calculate the impulse of the ball on the ball · ·

A

3.0Ns

B

4.5Ns

C

7.5Ns

D

12.0Ns

CORRECT OPTION:
b

Impulse = m(v-u)

= 0.15(80-50)

= 4.5N/s

= 0.15(80-50)

= 4.5N/s

7

The force of attraction between two point masses is 10^{-4}N when the distance between them is 0.18m. If the distance is reduced to 0.06m, calculate the force.

A

1.1 x 10^{-5}N

B

3.3 x 10^{-5}N

C

3.0 x 10^{-4}N

D

9.0 x 10^{-4}N

CORRECT OPTION:
d

F \(\alpha \frac{1}{r^2}\)

therefore, \(\frac{F_1}{F_2}\) = \(\frac{r_1}{(r_2)^2}\)

\(\frac{10^{-4}}{F_2}\) = \(\frac{(0.06)^2}{(0.18)^2}\)

F = 9 x 10^{-4}N

therefore, \(\frac{F_1}{F_2}\) = \(\frac{r_1}{(r_2)^2}\)

\(\frac{10^{-4}}{F_2}\) = \(\frac{(0.06)^2}{(0.18)^2}\)

F = 9 x 10

8

A uniform meter rule weighing 0.5N is to be pivoted on a knife-edge at the 30cm-mark. where will a force of 2N be placed from the pivot to balance the meter rule?

A

95cm

B

25cm

C

20cm

D

5cm

CORRECT OPTION:
b

9

A man whose mass is 80kg climbs a staircase in 20s and expends power of 120W. Find the height of the staircase.[g = 10ms^{-2}]

A

1.8m

B

2.0m

C

2.5m

D

3.0m

CORRECT OPTION:
d

Power x time = mgh

120 x 20 = 80 x 10 x h

2400 = 800h

making h the subject of formula,divide both side by 800

\(\frac{2400}{800}\) = h

h = 3m

120 x 20 = 80 x 10 x h

2400 = 800h

making h the subject of formula,divide both side by 800

\(\frac{2400}{800}\) = h

h = 3m

10

A parachute attains a terminal velocity when

A

its density is equal to the density of air

B

the viscous force of the air and the upthrust completely counteract its weight

C

it expands as a result of reduced external pressure

D

the viscous force of the air is equal to the sum of the weight and upthrust

CORRECT OPTION:
b

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