Find the rms velocity of its molecules in ms^-1
T₁= 27°C + 273 = 300 K, T₂ = 127°e + 273 = 400 K
rms speed of molecules is \({ v }_{ rms }=\sqrt { \cfrac { 3RT }{ M } } \)
So it depends only on temperature,
v_rms ∝ \(\sqrt { T } \)
\(\cfrac { { v }_{ 1 } }{ { v }_{ 2 } } =\sqrt { \cfrac { { T }_{ 1 } }{ { T }_{ 2 } } } \Rightarrow \cfrac { 200 }{ { V }_{ 2 } } =\sqrt { \cfrac { 300 }{ 400 } } ;\cfrac { 200 }{ { v }_{ 2 } } =\cfrac { \sqrt { 3 } }{ 2 } \)
\({ v }_{ 2 }=\cfrac { 400 }{ \sqrt { 3 } } \)
v₂ = 231ms^-1

According to acceleration due to gravity
\(\cfrac { g' }{ g } =\left( \cfrac { R }{ R+h } \right) ^{ 2 }\)
\(\cfrac { 1 }{ 2 } =\left( \cfrac { R }{ R+h } \right) ^{ 2 }\Rightarrow \cfrac { 1 }{ \sqrt { 2 } } =\cfrac { R }{ R+h } \)
\(R+h=2\sqrt { R } \Rightarrow h=(\sqrt { 2 } -1)R\)
h = 0.414R
Hence distance from centre = R + 0.414 R = 1.414 R

\({ v }_{ e }\)∝\(\cfrac { 1 }{ \sqrt { r } } \) where r is the position of body from the surface.
\(\cfrac { { v }_{ e } }{ v' } =\sqrt { \cfrac { R+7R }{ R } } =\sqrt { \cfrac { 8R }{ R } } =\sqrt { 8 } =2\sqrt { 2 } \)
\(\therefore v'=\cfrac { { v }_{ e } }{ 2\sqrt { 2 } } \)
 

This is because metals are good conductors of heat and have low specific heat capacity

\(\sum { { F }_{ y } } \)= ma_y=0
kx - mg = 0 => k =\(\frac{mg}{x}\)
The force constant is,
k=\(\frac { (1.5kg)(9.8m/{ s }^{ 2 }) }{ (0.020m) } =\frac { 14.7 }{ 0/020 } \) =735 N/m
Frequency =\(\frac { 1 }{ 2\pi } \sqrt { \frac { k }{ m } } \)
\(\Rightarrow\)4\(\pi ^{2}\)f²=\(\frac{k}{m}\)
m=\(\frac { k }{ 4{ \pi }^{ 2 }{ f }^{ 2 } } \Rightarrow m=\frac { (735N/m) }{ 4{ \pi }^{ 2 }({ 3.0Hz) }^{ 2 } } \)
=\(\frac { (735N/m) }{ 4\times ({ 3.75) }^{ 2 }\times ({ 3.0Hz) }^{ 2 } } \)
=\(\frac { 735N/m }{ 4\times 9.85\times 9.0Hz } \)
= \(\frac{735\ N/m}{354.6Hz}\) = 2.07 kg

The intermolecular forces are attractive, but they are repulsive for intermolecular reparations less than 10^-10m.

\( Y=2G(1+\sigma)\)
So, 2-4G = 2G(1+σ)
or σ =1.2-1= 0.2

Yes, for example in an adiabatic compression, temperature rises and in an adiabatic expansion, temperature falls, although no heat is given Dr taken from the system.

Brownian motion. In this motion any particular molecule will follow a zig - zag path due to be collision with the other molecule or with the walls of the container.

Phase difference = \(\frac{2\pi}{\lambda}\) (path difference)
\(\Delta \varphi =\frac { 2\pi }{ \lambda } \Delta r=\frac { \lambda }{ 2\pi } \Delta \varphi \)

For a fixed temperature, when the pressure varies, correspondingly density also varies such that the ratio \((\frac{p}{\rho})\) becomes constant. This means that the speed of sound is independent of pressure for a fixed temperature.

The displacement time t = 0s(initial time), the phase φ=φ0 is called epoch (initial phase) where φ0 is called the angle of epoch.

When a room has furniture, reverberation time can be suitably decreased, since furniture have large absorption coefficient of sound so, a tone sounds with lesser amplitude and intensity. Whereas in an empty room, reverberation time will be more than a room having things like furniture.

Any oscillation driven by an external periodic agency to overcome the damping is known as forced oscillation.
Example: Sound boards of stringed instruments.

Molar mass of nitrogen m = 0.028 kg/mol
Temperature T = 20 + 273 = 293 K
Universal gas constant \( R=8.314 \mathrm{~J} / \mathrm{mol} / \mathrm{k}\)
RMS speed of nitrogen (N₂),
\(v_{r m s} =\sqrt{\frac{3 R T}{m}}  \)
\(\left(\mathrm{~N}_{2}\right), v_{m s} =\sqrt{\frac{3 \times 8.31 \times 273}{0.028}}  \)
\(\left(\mathrm{~N}_{2}\right), v_{r m s} =\sqrt{2610 \times 10^{2}}=511 \mathrm{~m} / \mathrm{s}\)
Molar mass of Oxygen = 0.032 kg / mol
RMS speed of oxygen (O₂),
\(v_{m s}=\sqrt{\frac{3 R T}{m}}\)
RMS speed of (O₂) \(v_{m s} =\sqrt{\frac{3 \times 8.31 \times 293}{0.032}} \)
\(=\sqrt{2280 \times 10^{2}}\)
RMS speed of \(\left(\mathrm{O}_{2}\right), \quad v_{m s}=478 \mathrm{~m} / \mathrm{s}\)

The escape speed of gases on the surface of Moon is much less than the root mean square speeds of gases due to low gravity. Due to this all the gases escape from the surface of the Moon.

The pressure inside the balloon
\( P=1.8 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2} \)
The radius of the balloon \(R=10 \times 10^{-2} \mathrm{~m}\)
Room temperature \(T=(30+273)=303 \mathrm{~K}\)
Number of moles of air \(\mu=\frac{V P}{R T}\)
Universal gas constant \(\mathrm{R}=8.314 \mathrm{~J} / \mathrm{K} / \mathrm{mol}\)
Volume of air in balloon \(\mathrm{V}=\frac{4}{3} \pi R^{3}\)
\(\mathrm{V} =\frac{4}{3} \times 3.14 \times\left(10 \times 10^{-2}\right)^{3}  \)
\(=4.1866 \times 10^{-3} \mathrm{~m}^{3}  \)
\(\text {Number of moles of air } \mu =\frac{4.1866 \times 10^{-3} \times 1.8 \times 10^{5}}{8.314 \times 303}  \)
\(\mu  \simeq 0.3 \mathrm{moles}\)

When we suck through the straw, the pressure inside the straw becomes less than the atmospheric pressure. Due to the pressure different the soft drink rises in the straw and we are able to take the soft drink easily.

The weight of an object is defined as the downward force whose magnitude W is equal to the upward force that must be applied to the object to hold it at rest or at constant velocity relative to the Earth.

Total distance to be covered = 1 km + 100 m = 1100 m (including both bridge and time)
Then, Speed=60 kmh^-1\(=60\times\frac{5}{18}ms^{-1}=\frac{50}{3}\ ms^{-1}\)
Then, time taken to cover this distance \(=\frac{1100}{\frac{150}{9}}s=66s\)

1. An object dropped from window of a moving train.
2. A bullet fired from a rifle.
3. A ball thrown in any direction.
4. A javelin or shot put thrown by an athlete.
5. A jet of water issuing from a hole near the bottom of a water tank.

No change in speed of system as no external force is working.

Force \(\vec F=\vec i+2\vec j+3\vec k\)
Distance \(\vec S=4\vec i+6\vec j\)
Work done  \(\vec F.\vec S=(\vec i2\vec j+3\vec k).(4\vec i+6\vec j)=4+12+0=16J\)

The distance of the planet from the Earth
d = 6.3\(\times\)10¹⁰ m
Time t = 7 minutes = 7\(\times\)60 s.
The speed of signal v = ?
The speed of signal v = \({2d\over t}={2\times 6.3 \times 10^8\over 7\times 60}=3\times 10^8 ms^{-1}\)

The moment of inertia of the system (circular table + two boys) decreases. To conserve angular momentum (L = 100 = constant), the speed of rotation of the circular table increases.

When bodies have the same mass i.e., m₁ = m₂ and second body (usually called target) is at rest (u₂ = 0), By substituting m₁ = m₂ and u₂ = 0 in equations
\({ v }_{ 1 }=\left( \frac { { m }_{ 1 }-{ m }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } \right) { u }_{ 1 }+\left( \frac { { 2m }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } \right) { u }_{ 2 }\)...(1)
and equations \({ v }_{ 2 }=\left( \frac { { 2m }_{ 1 } }{ { m }_{ 1 }+{ m }_{ 2 } } \right) { u }_{ 1 }+\left( \frac { { m }_{ 2 }-{ m }_{ 1 } }{ { m }_{ 1 }+{ m }_{ 2 } } \right) { u }_{ 2 }\) , we get...(2)
from equations (1) => v₁ = 0
from equations (2) => v₂ = u₁

(i) Ball bearings provides another effective way to reduce the kinetic friction.
(ii) If ball bearings are fixed between two surfaces, during the relative motion only the rolling friction comes to effect and not kinetic friction.
(iii) The rolling friction is much smaller than kinetic friction; hence the machines are protected from wear and tear over the years.

Free body diagram is a simple tool to analyse the motion of the object using Newton's laws.
The following systematic steps are followed for developing the free body diagram:
(i) Identify the forces acting on the object.
(ii) Represent the object as a point.
(iii) Draw the vectors representing the forces acting on the object.

The K.E. of A without friction E_A = \(\frac{1}{2}\) m(2gh).
The K.E. of B undergoes pure rolling E_B
\(= \frac{1}{2}m \left( \frac { 2gh }{ 1+\frac { { K }^{ 2 } }{ { R }^{ 2 } } } \right) \)
The K.E. of C rolling with slipping E_c
=\(\frac{1}{2}\) m(2gh).

Lifting a table from the floor by two persons. Pushing the car by two persons.

(i) A uniform velocity in the horizontal direction, which will not change provided there is no air resistance.
(ii) A uniformly changing velocity (i.e., increasing or decreasing) in the vertical direction.

True

Power is a scalar quantity. Its dimension is [ML²T^-3]. The SI unit of power is watt (W).

Technology is the application of the principles of physics for practical purposes. The application of knowledge for practical purposes in various fields to invent and produce useful products or to solve problems is known as technology

Range of masses: from heavenly bodies to electron, 10⁵⁵ kg (mass of known observable universe) to 10^-31 kg (mass of an electron) [the actual mass of an electron is 9.11\(\times\)10^-31 Kg].

Length of a simple pendulum \(l=20 \mathrm{~cm}=20 \times 10^{-2} \mathrm{~m}\)
\(\text {Accuracy }=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}\)
Time for one oscillation \(=\frac{40}{50}=0.88\)
\(T =2 \pi \sqrt{\frac{l}{g}} ; T^{2}=4 \pi^{2}\left(\frac{l}{g}\right) \)
\(g =\frac{4 \pi^{2} l}{T^{2}}\)
\(\frac{\Delta g}{g} \times 100 ==\frac{\Delta l}{l} \times 100+2 \frac{\Delta T}{T} \times 100  \)
\(l =20 \mathrm{~cm} \Delta l=2 \mathrm{~mm}=0.2 \mathrm{~cm} \Delta T=1 \mathrm{~s}  \)
\(\frac{\Delta g}{g} =\frac{\Delta l}{l}+2 \frac{\Delta T}{T}  \)
\(=\frac{0.2}{20}+2 \times \frac{1}{40} \)
= 0.01 + 0.05 = 0.06
\(\therefore\) Percentage accuracy in the determination of g=0.06 x 100=6 %

Given
Function \(a \rightarrow y=x^{2}+2 \alpha t x\)
Function \(b \rightarrow y=(x+\mathrm{vt})^{2}\)
Formula
For the function to be a wave function
\(\frac{d y / d x}{d y / d t}\) must be a constant.
For function (a)
\(y =x^{2}+2 \alpha t x  \)
\(\frac{d y}{d x} =2 x+2 \alpha t  \)   .....(1)
\(\frac{d y}{d t} =0+2 \alpha x=2 \alpha x\)   .......(2)
Dividing equation (1) by (2) we get
\(\frac{d y / d x}{d y / d t}=\frac{2 x+2 \alpha t}{2 d x} \text { is not a constant }\)
∴ Function (a) is not describing wave
For function b :
\(\mathrm{y} =(x+\mathrm{vt})^{2}  \)
\(\therefore \frac{d y}{d x} =2(x+\mathrm{vt}) \times 1=2(x+\mathrm{vt}) \)   .....(3)
\(\frac{d y}{d t} =2(x+\mathrm{vt}) \times \mathrm{v} \)
\(=2 \mathrm{v}(x+\mathrm{vt})\)  ......(4)
Dividing equation (3) by equation (4) we get
\(\frac{d y / \mathrm{dx}}{d y / \mathrm{dt}}=\frac{2(x+v t)}{2 v(x+v t)}=\frac{1}{v}=\text { constant }\)
Hence function b satisfies the wave equation.

Time period of a simple pendulum is given by
T = \(2\pi \sqrt { \frac { l }{ g } } \)
l - length of a simple pendulum
g - acceleration due to gravity

Ratio of orbital distances \(\frac{d_{1}}{d_{2}}=2\).
Gravitational field \(\mathrm{E} =-\frac{G M}{r^{2}} \hat{r}  \)
\(\mathrm{E}_{1} =-\frac{G M}{d_{1}^{2}}  \)
\(\mathrm{E}_{2} =-\frac{G M}{d_{2}^{2}}  \)
\(\frac{E_{1}}{E_{2}} =-\frac{G M}{d_{1}^{2}} \times \frac{d_{2}^{2}}{(-G M)}  \)
\(\frac{E_{1}}{E_{2}} =\left(\frac{d_{2}}{d_{1}}\right)^{2}=\left(\frac{1}{2}\right)^{2}=\frac{1}{4}  \)
\(\therefore \mathrm{E}_{2} =4 \mathrm{E}_{1}\)

Mass m = 10 kg
Force F = 16 N
time t = 10 s
\(a=F/m=\frac { 16N }{ 10 \ kg } =1.6 \ ms^{ -2 }\)
We know that, v = u + at = 0 + 1.6 \(\times\) 10 = 16 ms^-1
Kinetic energy K.E = \(\frac { 1 }{ 2 } { mv }^{ 2 }=\frac { 1 }{ 2 } \times 10\times 16\times 16=1280\ J\)

It is a uniform motion

Distance travelled from 0 to 2 s= Area of the shaded portion
\(=\frac{1}{2} b_{1} h_{1}+\frac{1}{2} b_{2} h_{2}=\frac{1}{2} \times 1.5 \times 2+\frac{1}{2} \times 0.5 \times 1 \)
= 1.50 + 0.25 = 1.75m
Displacement
\(=-\frac{1}{2} b_{1} h_{1}+\frac{1}{2} b_{1} h_{2}  \)
\(=-\frac{1}{2} \times 1.5 \times 2+\frac{1}{2} \times 0.50 \times 1 \)
= -1.50 + 0.25 = -1.25 m
0-1 s The velocity increases in the negative direction
1 s-2 s Velocity increases in the positive direction. Change of velocity is 1-(-2)=3 ms^-1
2 s-5 s Velocity does not change. So there is no acceleration.
5 s-6 s Velocity reduces to zero.
6 s-7 s The body is at rest.