1

Which of the following is the most suitable for use as an altimeter?

A

A mecury barometer

B

A fortin barometer

C

A mecury manometer

D

An aneroid barometer

CORRECT OPTION:
d

2

A body of weight W N rest on a smooth plane inclined at an angle\(\theta\)^{o} to the horizontal. What is the resolved part of the weight in Newtons along the plane?

A

W sin\(\theta\)

B

W cos\(\theta\)

C

W sec\(\theta\)

D

W tan\(\theta\)

CORRECT OPTION:
a

horizontal component = W cos(90^{o} - \(\theta\))

= W sin^{}

= W sin

3

A small metal ball thrown vertically upwards from the top of a tower with an initial velocity of 20ms^{-1}. If the ball took a total of 6s to reach ground level, determine the height of the tower [g = 10ms^{-2}]

A

60m

B

80m

C

100m

D

120m

CORRECT OPTION:
a

Time from tower to top, t = \(\frac{u - v}{g}\)

= \(\frac{20 - 0}{10}\)

= 2s

distance from tower to top

x = ut - \(\frac{1}{2}\)gt^{2}

= (20 x 2) - (\(\frac{1}{2}\) x 10 x 2 x 2)

= 20m

Time from the top to the ground, t = 6 - 2

= 4s

distance from top to ground,

x = ut - \(\frac{1}{2}\)gt^{2}

= (0 x 4) + (\(\frac{1}{2}\) x 10 x 4 x 4)

= 80cm

therefore, height of tower = 80 - 20

= 60m

= \(\frac{20 - 0}{10}\)

= 2s

distance from tower to top

x = ut - \(\frac{1}{2}\)gt

= (20 x 2) - (\(\frac{1}{2}\) x 10 x 2 x 2)

= 20m

Time from the top to the ground, t = 6 - 2

= 4s

distance from top to ground,

x = ut - \(\frac{1}{2}\)gt

= (0 x 4) + (\(\frac{1}{2}\) x 10 x 4 x 4)

= 80cm

therefore, height of tower = 80 - 20

= 60m

4

An object moves with uniform speed round a circle. Its acceleration has

A

constant magnitude and constant direction

B

constant magnitude and varying direction

C

varying magnitude but constant direction

D

varying magnitude and varying direction

CORRECT OPTION:
b

5

A body of mass 100g moving with a velocity of 10.0ms^{-1} collides with a wall. If after collision, it moves with a velocity of 2.0ms^{-1} in the opposite direction, calculate the change in momentum.

A

0.8Ns

B

1.2Ns

C

12.0Ns

D

80.0Ns

CORRECT OPTION:
b

Initial momentum = 0.1 x 10 = 1Ns

final velocity = -2ms^{-1}

Final momentum = -2 x 0.1 = -0.2Ns

Change in momentum = -0.2 - 1 = -1.2Ns

final velocity = -2ms

Final momentum = -2 x 0.1 = -0.2Ns

Change in momentum = -0.2 - 1 = -1.2Ns

6

A spring of force constant 1500Nm^{-1} is acted upon by a constant force of 75N. Calculate the potential energy stored in the string

A

1.9J

B

2.8J

C

3.45J

D

0.43J

CORRECT OPTION:
a

P.E = \(\frac{1}{2}\)Fe

= \(\frac{1}{2}\)\(\frac{F}{K}\)

= \(\frac{1}{2}\) x \(\frac{75 \times 75}{1500}\)

= 1.9J

= \(\frac{1}{2}\)\(\frac{F}{K}\)

= \(\frac{1}{2}\) x \(\frac{75 \times 75}{1500}\)

= 1.9J

7

A wheel and axle have radii 80cm and 10cm respectively. If the efficiency of the machine is 0.85, an applied force of 1200N to the wheel will raise a load of

A

8.0N

B

6.8N

C

8160N

D

9600N

CORRECT OPTION:
c

Efficiency = m.A x \(\frac{1}{V.R}\) x 100%

85 = \(\frac{load}{1200}\) x \(\frac{10}{80}\) x \(\frac{100}{1}\)

load = 8160N

85 = \(\frac{load}{1200}\) x \(\frac{10}{80}\) x \(\frac{100}{1}\)

load = 8160N

8

A 20kg mass is to be pulled up a slope inclined at 30^{o} to the horizontal. If the efficiency of the plane is 75%, the force required to pull the load up the plane is [g = 10ms^{-2}]

A

13.3N

B

73.5N

C

133.3N

D

53.2N

CORRECT OPTION:
c

Efficiency = m.A x \(\frac{1}{V.R}\) X 100%

75 = \(\frac{20 \times 10}{effort}\) x \(\frac{1}{\frac{g}{sin\theta}}\) x 100%

Effort = 133.3N

75 = \(\frac{20 \times 10}{effort}\) x \(\frac{1}{\frac{g}{sin\theta}}\) x 100%

Effort = 133.3N

9

The spiral spring of a spring balance is 25.0cm long when 5N hangs on it and 30cm long, when the weight is 10N. What is the length of the spring if the weight is 3N assuming Hooke's law is obeyed?

A

15.0cm

B

17.0cm

C

20.0cm

D

23.0cm

CORRECT OPTION:
d

let the length be Lcm

\(\frac{5 - 3}{25 - L}\) = \(\frac{10 - 5}{30 - 25}\)

L = 23cm

\(\frac{5 - 3}{25 - L}\) = \(\frac{10 - 5}{30 - 25}\)

L = 23cm

10

The mass of a stone is 15.0g when completely immersed in water and 10.0g when completely immersed in a liquid of relative density 2.0. The mass of the stone in air is

A

5.0g

B

12.0g

C

20.0g

D

25.0g

CORRECT OPTION:
c

R.D = \(\frac{\text{ mass of liquid displaced}}{\text {mass of water displaced}}\)

2 = \(\frac{m - 10}{m - 15}\)

m = 20g

2 = \(\frac{m - 10}{m - 15}\)

m = 20g

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