Mass spectrograph.

No, given \(\underset { i }{ \Sigma } { F }_{ i }\neq 0\)
The sum of torques about a certain point O
\(\underset { i }{ \Sigma } { r }_{ i }\times F_{ i }=0\)
The sum of torques about any other point O'.
\(\underset { i }{ \Sigma } ({ r }_{ i }-a)\times F_{ i }=\underset { i }{ \Sigma } { r }_{ i }\times F_{ i }-a\times \underset { i }{ \Sigma } { F }_{ i }\)
Here, the second term need not vanish.

It is normal to the plane containing the earth and the sun as areal velocity. \(\frac { \triangle A }{ \triangle t } =\frac { 1 }{ 2 } r\times v\triangle t\) and directed according to right hand rule.

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.

When air is pumped, more molecules are pumped and Boyle's law is stated for situation where number of molecules constant.

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

Propagation of longitudinal waves through a medium leads to transmission of energy through the medium.

Speed of sound waves in air increases with increases with increase in humidity. This is because presence of moisture decreases the density of air.

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.
\(\Rightarrow v^{ , }=v\Rightarrow \frac { 2v }{ \lambda ^{ ' } } =\frac { v }{ \lambda } \Rightarrow { \lambda }^{ ' }=2\lambda \)

Given, v=400 Hz, V_s = 10m/s, v = 330m/s
As the source is moving towards stationary observer, apparent frequency will be more than the original.
V^'=apparent frequency
\(=\frac { v\times V }{ v-{ v }_{ s } } =\frac { 330\times 400 }{ 330-10 } \)
\(=412.5\ Hz\)

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.