The sonometer frequency is given by
v=\(\frac{n}{2L}\sqrt{\frac{T}{μ}}\)
Now, as it vibrates with length L, we assume v=v₁
n = n₁
∴v₁=\(\frac{n_1}{2L}\sqrt{\frac{T}{μ}}\)
When length is doubled, then
v₂=\(\frac{n_2}{2 \times2L}\sqrt{\frac{T}{μ}}\)
Dividing Eq.(i) by Eq. (ii), we get
\(\frac{v_1}{v_2}=\frac{n_1}{n_2}×2\)
To keep the resonance,
\(\frac{v_1}{v_2}=1=\frac{n_1}{n_2}×2\)
⇒n₂=2n₁
Hence, when the wire is doubled, the number of loops also get doubled to produce the resonance. That is, it resonates in second harmonic.

Frequency in the first medium, v =v/λ
Frequency will remain same in the second medium, as the source.
⇒v, =v ⇒ 2v/λ′ = v/λ ⇒ λ′ = 2λ

The resultant amplitude will be
y=\(\sqrt{y_2^1+y_2^2}=\sqrt{9+16}=\sqrt{25}=\)5cm

Water waves produced by a motorboat sailing in water are both longitudinal and transverse.

Total distance travelled by an oscillator in one time period, from its mean position to one extreme position, then to other extreme position and finally back to mean position is 4 A, where A is the amplitude of oscillation.
Hence, the ratio =4A/A=4

When the displacement amplitude of the pendulum is extremely small as compared to its length.

As, we know, mean free path, 
λ ∝ 1/d²
Given, d₁= 1A^o and d₂ = 2A^o⇒ λ₁ : λ₂=4:1

Keeping p constant, we have
V²/2 = \(\frac{V_1T_2}{T_1}\)
= \(\frac{100×600}{300}\)
= 200cc

Here, heat removed is less than the heat supplied and hence the room become hotter.

During driving, temperature of the gas increases while its volume remains constant. So, according to Charles' law, at constant V, p ∝  T. Therefore, pressure of gas increases

If the system does work against the surroundings so that it compensates for the heat supplied, the temperature can remain constant.

No, because stress is a scalar quantity, not a vector quantity.
Stress = Magnitude of internal reaction force/Area of cross−section 

At perihelion, because the earth has to cover greater linear distance to keep the areal velocity constant.

Suppose, b be the position vector of centre of mass of regular n-polygon.
As (n-1) equal point masses each of mass m are placed at (n-1) vertices of regular polygon, therefore
\( \frac{(n-1) m b+m a}{(n-1+1) m}=0 \)
\( \Rightarrow(n-1) m b+m a=0 \)
\( \Rightarrow b=\frac{-a}{(n-1)} \)

It least count of physical balance is 0.1 g, the mass measured will be correctly represented as m=(156.3±0.1)g