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A force of 50 dynes is acted on a body of mass 5 g which is at rest for an interval of 3 seconds, then impulse is

$0 .15 \times 10^{- 3} N s$

$0 .98 \times 10^{- 3} N s$

$1 .5 \times 10^{- 3} N s$

$2 .5 \times 10^{- 3} N s$

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Explanation

Impulse, $I = F \times Δ t = 50 \times 10^{- 5} \times 3$ = $1 .5 \times 10^{- 3} N - s$

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A bullet of mass 0.1 kg is fired with a speed of 100 m/sec, the mass of gun is 50 kg. The velocity of recoil is

0.2 m/sec

0.1 m/sec

0.5 m/sec

0.05 m/sec

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Explanation

According to principle of conservation of linear momentum, $m_{G} v_{G} = m_{B} v_{B}$

$\Rightarrow v_{G} = \frac{m_{B} v_{B}}{m_{G}} = \frac{0 .1 \times 10^{2}}{50} = 0 .2 m / s$

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Two forces of magnitude F have a resultant of the same magnitude F. The angle between the two forces is

45°

120°

150°

60°

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Explanation

$F = \sqrt{{(F)}^{2} + {(F)}^{2} + 2 F . F \cos θ}$ ⇒ $θ = 120 °$

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Two forces with equal magnitudes F act on a body and the magnitude of the resultant force is F/3. The angle between the two forces is

$\cos^{- 1} (- \frac{17}{18})$

$\cos^{- 1} (- \frac{1}{3})$

$\cos^{- 1} (\frac{2}{3})$

$\cos^{- 1} (\frac{8}{9})$

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Explanation

$F_{_{n e t}}^{2} = F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} \cos θ$

⇒ ${(\frac{F}{3})}^{2} = F^{2} + F^{2} + 2 F^{2} \cos θ$

⇒ $\cos θ = (- \frac{17}{18})$

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If two forces of 5 N each are acting along X and Y axes, then the magnitude and direction of resultant is

$5 \sqrt{2}, π / 3$

$5 \sqrt{2}, π / 4$

$- 5 \sqrt{2}, π / 3$

$- 5 \sqrt{2}, π / 4$

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Explanation

$| \vec{F} | = \sqrt{5^{2} + 5^{2}} = 5 \sqrt{2} N .$

and $\tan θ = \frac{5}{5}$ =1

$\Rightarrow θ = π / 4 .$

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A block is kept on a frictionless inclined surface with angle of inclination 'α'. The incline is given an acceleration 'a' to keep the block stationary. Then a is equal to

g tan α

g / tan α

g cosec α

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Explanation

For a block to remain stationary on a frictionless inclined surface with angle Î±, the acceleration a given to the incline must be equal to g tan Î±, where g is the acceleration due to gravity. This acceleration provides just enough force component along the incline to balance the weight component.

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Two masses 2 kg and 3 kg are attached to the end of the string passed over a pulley fixed at the top. The tension and acceleration are

$\frac{7 g}{8}; \frac{g}{8}$

$\frac{21 g}{8}; \frac{g}{8}$

$\frac{21 g}{8}; \frac{g}{5}$

$\frac{12 g}{5}; \frac{g}{5}$

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Explanation

$T = \frac{2 m_{1} m_{2}}{m_{1} + m_{2}} g = \frac{2 \times 2 \times 3}{2 + 3} g = \frac{12}{5} g$

$a = (\frac{m_{2} - m_{1}}{m_{1} + m_{2}}) g = (\frac{3 - 2}{3 + 2}) g = \frac{g}{5}$

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Two bodies of mass 3 kg and 4 kg are suspended at the ends of massless string passing over a frictionless pulley. The acceleration of the system is (g = 9.8 m/s²)

4.9 m/s²

2.45 m/s²

1.4 m/s²

9.5 m/s²

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Explanation

Acceleration $= \frac{(m_{2} - m_{1})}{(m_{2} + m_{1})} g$

$= \frac{4 - 3}{4 + 3} \times 9 .8 = \frac{9 .8}{7} = 1 .4 m / \sec^{2}$

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Three solids of masses m₁, m₂ and m₃ are connected with weightless string in succession and are placed on a frictionless table. If the mass m₃ is dragged with a force T, the tension in the string between m₂ and m₃ is

$\frac{m_{2}}{m_{1} + m_{2} + m_{3}} T$

$\frac{m_{3}}{m_{1} + m_{2} + m_{3}} T$

$\frac{m_{1} + m_{2}}{m_{1} + m_{2} + m_{3}} T$

$\frac{m_{2} + m_{3}}{m_{1} + m_{2} + m_{3}} T$

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The mass of a body measured by a physical balance in a lift at rest is found to be m. If the lift is going up with an acceleration a, its mass will be measured as

$m (1 - \frac{a}{g})$

$m (1 + \frac{a}{g})$

Zero

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Explanation

Mass measured by physical balance remains unaffected due to variation in acceleration due to gravity.

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