This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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Step 1: Solve Question 1. The initial velocity is u = 0 m/s (starts from rest). The final velocity is v = 72 km/h. Convert this to m/s: v = 72 km/h × 1000 m1 km × 1 h3600 s = (72000)/(3600) m/s = 20 m/s The time taken is t = 10 s. Using the first equation of motion, v = u + at: 20 m/s = 0 m/s + a × 10 s a = 20 m/s10 s = 2 m/s^2 The correct option is b. b) 2 m/s^2 Step 2: Solve Question 2. Kinetic energy (KE) is given by the formula KE = (1)/(2)mv^2, where m is mass and v is velocity. For two identical bodies, their masses (m) are the same. If one body has a higher velocity, its kinetic energy will be higher because KE is directly proportional to the square of the velocity. The correct option is b. b) Higher Kinetic Energy Step 3: Solve Question 3. Assertion: A table cloth cannot be pulled from a table without dislodging the dishes. This assertion is generally false. The "tablecloth trick" works because the dishes, due to their inertia, tend to resist the change in motion when the cloth is pulled quickly. Reason: Newton's second law of motion gives definition of inertia. This reason is false. Newton's first law of motion defines inertia. Since both the assertion and the reason are false, the correct evaluation would be that both are incorrect. Step 4: Solve Question 4. The electrical energy consumed is E = 1000 J. The time taken is t = 10 s. Power (P) is defined as energy consumed per unit time: P = (E)/(t) P = 1000 J10 s = 100 W 100 W Step 5: Solve Question 5. The distance to the wall is d = 55 m. The speed of sound is v = 330 m/s. For an echo to be heard, the sound travels to the wall and then reflects back to the listener. So, the total distance covered by the sound is 2d. Total distance = 2 × 55 m = 110 m. The time (t) taken for the echo is: t = Total distanceSpeed of sound t = 110 m330 m/s = (1)/(3) s ≈ 0.333 s 0.333 s Step 6: Solve Question 5 (OR part). What is the work done by the force of gravity on a satellite moving round the Earth? Justify your answer. A satellite moving in a circular orbit around the Earth experiences a gravitational force that is always directed towards the center of the Earth. The displacement of the satellite at any instant is tangential to its circular path. The angle () between the force of gravity and the direction of displacement is 90^. The work done (W) by a force is given by the formula W = Fd . Since = 90^, 90^ = 0. Therefore, the work done by the force of gravity on a satellite moving in a circular orbit is zero. Work done is 0 J Justification: The force of gravity acts perpendicular to the displacement of the satellite at every point in its circular orbit. Step 7: Solve Question 6. Define the characteristic of sound. (also write unit). i) Amplitude: The maximum displacement or distance moved by a point on a vibrating body or wave measured from its equilibrium position. It determines the loudness or intensity of the sound. Unit: meter (m) for displacement, or Pascal (Pa) for sound pressure. ii) Pitch: The perception of the frequency of a sound. A higher frequency corresponds to a higher pitch, and a lower frequency corresponds to a lower pitch. Unit: Hertz (Hz) (as it is directly related to frequency). iii) Frequency: The number of complete oscillations or cycles of a wave that occur per unit of time. It determines the pitch of the sound. Unit: Hertz (Hz). Step 8: Solve Question 7. Give reason for the following. i) Some of the leaves may get detached from a tree if we vigorously shake its branch. This phenomenon is due to inertia of rest. When the branch is shaken vigorously, it moves suddenly. However, the leaves, due to their inertia, tend to remain in their original state of rest. This resistance to change in motion causes the leaves to detach from the moving branch. ii) While catching a fast moving cricket ball, a fielder in the ground gradually pulls his hands backwards with the moving ball. This action is performed to increase the time interval ( t) over which the momentum of the ball changes. According to the impulse-momentum theorem (F t = p), if the change in momentum ( p) is constant, increasing the time interval ( t) over which the force acts will decrease the magnitude of the force (F) exerted on the fielder's hands. This reduces the impact force and prevents injury.