Assertion (A): In a stationary-wave system, displacement nodes are pressure antinodes, and displacement antinodes are pressure nodes.
Reason (R): When a closed organ pipe vibrates, the pressure of the gas at the closed end remains constant.
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
In a stationary wave, displacement nodes correspond to points of maximum pressure variation (pressure antinodes) and displacement antinodes correspond to points of minimum pressure variation (pressure nodes). So, (A) is true.
At the closed end of an organ pipe, there must be a displacement node, which is a pressure antinode, meaning there is maximum pressure variation, not constant pressure. Thus, (R) is false.
Assertion (A): Both arms of a tuning fork vibrate with the same frequency.
Reason (R): The two arms of a tuning fork vibrate in phase.
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
Both arms of a tuning fork vibrate as a single system, thus at the same frequency. Assertion (A) is true.
For efficient sound production, the arms vibrate 180 degrees out of phase (in opposite directions), not in phase. Therefore, Reason (R) is false.
Assertion (A): The energy stored by a stationary wave is zero.
Reason (R): When two identical waves travelling in opposite directions superimpose, their whole energy is converted into heat.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
Assertion (A) is false; a stationary wave stores energy, which oscillates between kinetic and potential forms.
Reason (R) is false; energy is not converted to heat but redistributed to form the stationary wave pattern.
Assertion (A): In \(n^{\text{th}}\) normal mode of a stretched string, there are \(n\) antinodes and \((n+1)\) nodes.
Reason (R): The ends of string are nodes, so the number of nodes is one more than the number of antinodes.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
Assertion (A) is true. For a stretched string fixed at both ends, the \(n^{\text{th}}\) normal mode indeed has \(n\) antinodes and \(n+1\) nodes.
Reason (R) is true and correctly explains this: the fixed ends are always nodes, leading to one more node than antinode.
Assertion (A): A tuning fork is in resonance with a closed pipe in fundamental mode, but the same tuning fork cannot be in resonance in fundamental mode with an open pipe of same length.
Reason (R): The same tuning fork will not be in resonance with open pipe of same length due to end correction of pipe.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
Assertion (A) is true. For a closed pipe of length \(L\), fundamental frequency is \(v/(4L)\). For an open pipe of same length, it's \(v/(2L)\). These are different, so the same tuning fork cannot resonate in fundamental mode with both. Reason (R) is false; the primary reason is the fundamental frequency difference (\(f_o = 2f_c\)), not solely end correction.