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

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
\(\frac { (n-1)mb+ma }{ (n-1+1)m } =0\)
\(\\ \Rightarrow (n-1)mb+ma=0\)
\(\\ \Rightarrow b=\frac { -a }{ (n-1) } \)

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

No, because stress is a scalar quantity, not a vector quantity.
Stress = \(\frac { Magnitude \ of \ internal \ reaction \ force }{ Area \ of \ cross-section } \) 

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

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

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

Keeping p constant, we have
\({ V }_{ 2 }\) = \(\frac { { V }_{ 1 }{ T }_{ 2 } }{ { T }_{ 1 } }\)
= \(\frac { 100\times 600 }{ 300 } \)
= 200cc

As, we know, mean free path, 
\(\lambda \propto \frac { 1 }{ { d }^{ 2 } } \)
\(Given,\ { d }_{ 1 }=1\overset { 0 }{ A } \ and\ { d }_{ 2 }=2\overset { 0 }{ A } \Rightarrow { \lambda }_{ 1 }:{ \lambda }_{ 2 }=4:1\)

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

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 \(=\frac { 4A }{ A } =4\)

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

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

Frequency in the first medium, v =\(\frac { v }{ \lambda } \)
Frequency will remain same in the second medium, as the source.
\(\Rightarrow v^{ , }=v\Rightarrow \frac { 2v }{ \lambda ^{ ' } } =\frac { v }{ \lambda } \Rightarrow { \lambda }^{ ' }=2\lambda \)

The sonometer frequency is given by
\(v=\frac { n }{ 2L } \sqrt { \frac { T }{ \mu } } \)
Now, as it vibrates with length L, we assume v=v₁
n = n₁
\(\therefore { v }_{ 1 }=\frac { { n }_{ 1 } }{ 2L } \sqrt { \frac { T }{ \mu } } \)
When length is doubled, then
\({ v }_{ 2 }=\frac { { n }_{ 2 } }{ 2\times 2L } \sqrt { \frac { T }{ \mu } } \)
Dividing Eq.(i) by Eq. (ii), we get
\(\frac { { v }_{ 1 } }{ { v }_{ 2 } } =\frac { { n }_{ 1 } }{ { n }_{ 2 } } \times 2\)
To keep the resonance,
\( \frac { { v }_{ 1 } }{ { v }_{ 2 } } =1=\frac { { n }_{ 1 } }{ { n }_{ 2 } } \times 2\)
\(\\ \Rightarrow { n }_{ 2 }=2{ n }_{ 1 }\)
Hence, when the wire is doubled, the number of loops also get doubled to produce the resonance. That is, it resonates in second harmonic.