Natural frequency of a body depends upon
(i) elastic properties of the material of the body and
(ii) dimensions of the body.

For rotational equilibrium of a body the vector sum of torques of all the forces acting on the body about the reference point must be zero.

It will move up.

At low temperature, diatomic gas has three translational and two rotational degrees of freedom, so a total number of degrees of freedom is 5.
But at high temp, gas molecule starts to vibrate which give two additional degrees of freedom i.e. 7.

For 1 mole with f degrees of freedom,
Internal energy, U = 1 x C_v x T = f²/RT
For n moles, U = nC_vT = nf²/RT

According to the principal of heat exchanges, all the objects (above 0 K) not only radiate electromagnetic energy but also absorb at the same rate from their surroundings. Thus they do not become colder.

Pressure is an external force per unit area, while stress is the internal restoring force which comes into play in  a deformed body acting transversely per unit area of  body.

Fiven F = 1N, m = 1 g  = 10^-3 kg
Now, F = ma ⟹ a = \(\frac{F}{m} = \frac{1}{10^{-3}}\)
= 10³ m/s²

No, because it is propelled by combustion of fuel and does not move under the effect of gravity alone.

Yes, Multiplying any two scalars is meaningful. Density p and volume V both the scalar quantities.When density is multiplied by volume,then we get \(p\times VB=m\)   , mass of the body,which is scalar quantity

Displacement travelled in time
(t₁ + t₂) = s₁+s₂
= v₁t₁ + v₂t₂
Average velocity \(=\frac { Net \ displacement }{ Total \ time \ taken } =\frac { { v }_{ 1 }{ t }_{ 1 }+{ v }_{ 2 }{ t }_{ 2 } }{ ({ t }_{ 1 }+{ t }_{ 2 }) } \)

Yes, when two bodies move in opposite direction then relative velocity of each is greater than the individual velocities.

Relative error = \(\frac { \triangle \bar { d } }{ d } =\frac { 0.01 }{ 0.39 } =0.0256\)

\([a]=[F]=\left[M L T^{-2}\right]\)
\([b]=\left[\frac{F}{x}\right]=\left[\frac{M L T^{-2}}{L}\right]=\left[M T^{-2}\right]\)

It is true that no physicist has ever seen an electron but several phenomena taking place in our daily life give us evidence of the existence of the electron such as frictional electricity.
On the other hand, no one has ever seen ghosts but there is no phenomena which can be explained on the basis of the existence of ghosts. Therefore, there is no comparison between two cases.

When no external force acts on a system the momentum will remain conserved. Consider a system of a n bodies of masses m₁, m₂, m₃ ....., m_n· If P_1' P_2' P_3' ....., P_n are the momentum associated then, the rate of change of momentum with the system,
\({dp\over dt}={dp_1\over dt}+{dp_2\over dt}+{dp_3\over dt}+....+{dp_n\over dt}\)
\({d\over dt}(p_1+p_2+p_3+.....+p_n)\)
If no external force acts,\({dp\over dt}=0\)
\(\therefore p=\) constant, i.e., \((p_1+p_2+p_3+.....+p_n)\) = constant.​​​​​​​​​​​​​

When the distance between two loudspeakers from the position of listener is an odd multiple of \(\frac{\lambda}{2}\), then due to destructive interference between sound waves from two loudspeakers, a feeble sound is heard by the listener.
When one loudspeaker is kept off, no interference is will take place and the listener will hear the full sound of the operating loudspeaker.

This is because a gas can be heated under different conditions of pressure and volume. The amount of heat required to raise the temperature of unit mass through unit degree is different under different conditions of heating

Here, Q₂ = 250 Js^-1
T₂ = -23^o C = -23 + 273 = 250 K
T₁ = 25^o C = 25 + 273 = 298 K
We know, \(\beta =\frac { { Q }_{ 2 } }{ W } =\frac { { T }_{ 2 } }{ { T }_{ 1 }-{ T }_{ 2 } } \)
\(W=\frac { { Q }_{ 2 }({ T }_{ 1 }-{ T }_{ 2 }) }{ { T }_{ 2 } } \)
\(\\ W=\frac { 250(298-250) }{ 250 } =\frac { 250\times48 }{ 250 } \)
\(\\ W=48{ Js }^{ -1 }\)

Conservation of energy gives
\( \frac{1}{2} m v^2-\frac{G M m}{R}=0-\frac{G M m}{(R+10 R)} \)
\(v=\sqrt{\left(\frac{20}{11} \frac{G m}{R}\right)=1.07 \times 10^4 m s^{-1}}\)

At maximum compression the kinetic energy of the car is converted entirely into the potential energy of the spring.
The kinetic energy of the moving car is
\(K=\frac{1}{2} m v^{2}\)
\(=\frac{1}{2} \times 10^{3} \times 5 \times 5\)
K = 1.25 x 10⁴ J
where we have converted 18 km h^–1 to 5 m s^–1 [It is useful to remember that 36 km h^–1 = 10 m s^–1]. At maximum compression x_m , the potential energy V of the spring is equal to the kinetic energy K of the moving car from the principle of conservation of mechanical energy.
\(V=\frac{1}{2} k x_{m}^{2}\)
= 1.25 x 10⁴ J
We obtain
x_m = 2.00 m
We note that we have idealised the situation. The spring is considered to be massless. The surface has been considered to possess negligible friction.

A vector, in general, has no definite location in space because a vector remains unaffected whenever it is displaced anywhere in space provided its magnitude and direction do not change. However, a position vector has a definite location in space.

We can add and structure only like quantities.
\(\Rightarrow\) Dimensions of P =  Dimensions of \(\frac{\mathrm{a}}{\mathrm{V}^2}\)
and dimension of V = Dimensions of b
From (i),
Dimensions of a = Dimensions of P x Dimensions of \(\mathrm{V}^2[\mathrm{a}]\)
\( =\left[\mathrm{M}^1 \mathrm{~L}^{-1} \mathrm{~T}^{-2}\right] \times\left[\mathrm{L}^3\right]^2 \)
\( =\left[\mathrm{M}^1 \mathrm{~L}^5 \mathrm{~T}^{-2}\right]\)
Unit of a = Unit of p x Unit of v²
\(=\frac{\mathrm{N}}{\mathrm{m}^2} \times \mathrm{m}^6=\mathrm{Nm}^4\)
From (ii), \([\mathrm{b}]=[\mathrm{V}]=\left[\mathrm{M}^0 \mathrm{~L}^3 \mathrm{~T}^0\right]\)
So unit of b = Unit of \(\mathrm{V}=\mathrm{m}^3\)
a = [M¹L⁵T^-2], b = [L³]

(a) Inside a liquid drop
Let r = radius of a spherieal liquid drop of centre O. T = surface tension of the liquid. Let P_i and P_o be the values of pressure inside and outside the drop.
\(\therefore\) Excess pressure inside the liquid drop = P_i - P_o
Let \(\Delta\)r be the increase in its radius due to excess pressure. It has one free surface outside.
\(\therefore\)  increase in surface area of the liquid drop
= \(4\pi(r+\Delta r)^2-4\pi r^2\)
= \(4\pi[r^2+(\Delta r)^2+2r\Delta r-r^2]\)
= \(8\pi r\quad \Delta r\)  .................(i)
\(\therefore\) increase in surface energy of the drop is
W = Surface tension x increase in area 
= \(T\times 8\pi r\quad \Delta r\)  .............. (ii)
Also W = Force due to excess of pressure x displacement
= Excess pressure x Area of drop x increase in radius
= \((p_i-p_0)4\pi r^2\Delta r\) ............(iii)
\(\therefore\) From eqns (ii) and (iii), we get
\((p_i-p_0)\times 4\pi r^2\quad \Delta r=T\times 8\pi r\quad \Delta r\)
\(\Rightarrow p_i-p_0=\frac{2T}{r}\)
(b) Inside a liquid bubbles:
A liquid bubble has air both inside and outside it and therefore it has two free surfaces.
Thus increase in its surfaces area
= \(2[4\pi (r+\Delta r)^2-4\pi r^2]\)
= \(2\times 8\pi r\Delta r=16\pi r\quad\Delta r\)
\(\therefore\)   W = \(T\times 16\pi r\Delta r\)     ...............(i)
Also  W = \((p_i-p_0)4\pi r^2\times \Delta r\)  ........(ii)
From eqnd (i) and (ii), we get
\((p_i-p_0)\times 4\pi r^2\times \Delta r=T.16\pi r\quad \Delta r\)
or \((p_i-p_0)=\frac{4T}{r}\)
(c) Inside an air bubble:
Air bubble is formed inside liquid, thus air bubble has one free surface inside it and liquid is outside.
If   r = radius of air bubble
\(\Delta\)r = increase in its radius due to excess of pressure
(P_i - P₀) inside it.
T = surface tension of the liquid in which bubble is formed.
\(\therefore\) increase in surface area = \(8\pi r \Delta r\).
\(\therefore\)    W = T x \(8\pi r \Delta r\)
Also W = \((p_i-p_0)\times 4\pi r^2\Delta r\)
\(\therefore (p_i-p_0)\times 4\pi r^2\Delta r=T\times 8\pi r^2\Delta r\)
or  \(p_i-p_0=\frac{2T}{r}\)

Consider an air chamber of volume V with a long neck of uniform area of cross-section A, and a frictionless ball of mass m fitted smoothly in the neck at position C, Fig. The pressure of air below the ball inside the chamber is equal to the atmospheric pressure. Increase the pressure on the ball by a little amount p. so that the ball is depressed to position D, where CD = y.
There will be decrease in volume and hence increase in pressure of air inside the chamber. The decrease in volume of the air inside the chamber, ΔV = Ay
Volumetric strain =\(\frac { change\ in\ volume }{ original\ volume } \)
\(=\frac { \Delta V }{ V } =\frac { Ay }{ V } \)
∴Bulk Modulus of elasticity E. will be
\(E=\frac { stress(or\ increase\ in\ pressure) }{ volumetric\ strain } \)
\(=\frac { -p }{ Ay/V } =\frac { -pV }{ Ay } \)
Here, negative sign shows that the increase in pressure will decrease the volume of air in the chamber.
Now, \(p=\frac { -EAy }{ V } \)
Due to this excess pressure, the restoring force acting on the ball is
\(F=p\times A=\frac { \_ EAy }{ V } .A=\frac { -E{ A }^{ 2 } }{ V } y\quad ...(i)\)
Since F ∝ y and negative sign shows that the force is directed towards equilibrium position. If the applied increased pressure is removed from the ball, the ball will start executing linear SHM in the neck of chamber with C as mean position.
In S.H.M., the restoring force,
F = -ky ...(ii)
Comparing (i) and (ii), we have
Spring factor, k = EA²/V
Here inertia factor mass of ball = m.
Period, T = \(2\pi \sqrt { \frac { inertia\ factor }{ spring\ factor } } \)
\(=2\pi \sqrt { \frac { m }{ { EA }^{ 2 }/V } } =\frac { 2\pi }{ A } \sqrt { \frac { mV }{ E } } \)
∴ Frequency, v = \(\frac { 1 }{ T } =\frac { A }{ 2\pi } \sqrt { \frac { E }{ mV } } \) 

When the tube is held horizontally, the mercury thread of length 76 cm traps a length of air = 15 cm. A length of 9 Cm of the tube will be left at the open end. The pressure of air enclosed in
tube will be atmospheric pressure. Let area of cross-section of the tube be 1sq. cm.
\(\therefore \) P₁ = 76 Cm  and V₁ = 15 Cm³
When the tube is held vertical1y, 15 em air gets another 9 cm of air (filled in the right handside in the horizontal position) and let h cm of mercury flows out to balance the atmospheric pressure. Then the heights of air column and mercury column are (24 + h) cm and (76 - h) cm respectively.
The pressure of air  = 76 - (76 - h) -h cm of mercury.
\(\therefore \)  V₂ = (24 + h) CM³  and P₂ = h cm
If we assume that temperature remains constant, then
P₁V₁ =P₂ V₂ or 76 x 15 = h x (24 + h) or h² + 24h - 1140 = 0
\(or \ h=\frac { -24\pm \sqrt { \left( { 24 } \right) ^{ 2 }+4\times 1140 } }{ 2 } =23.8\ cm\ or\ -47.8cm\)
Since h cannot be negative(because more mercury cannot flow into the tube), therefore h = 23.8 cm thus in the vertical position of the tube, 23.8 cm of mercury flows out.

We know that angular momentum \(\vec { l } \) of a particle having position vector \(\vec { r } \) and momentum \(\vec { p } \) is given by
\(\vec { l } \) = \(\vec { r } \)\(\times\)\(\vec { p } \)
But, \(\vec { r } \)= [x\(\vec { i} \)+ y\(\vec { j} \)+ z\(\vec { k} \)] where x,y,z are the components of
\(\vec { r } \)and\(\vec { p } \) = [p_x\(\vec { i} \)+ p_y\(\vec { j} \)+ p_z\(\vec { k} \)]
∴ \(\vec { l } \) = \(\vec { r } \)\(\times\)\(\vec { p } \) [x\(\vec { i} \)+ y\(\vec { j} \)+ z\(\vec { k} \)]\(\times\)[p_x\(\vec { i} \)+ p_y\(\vec { j} \)+ p_z\(\vec { k} \)]
or (lx\(\vec { i} \)+ ly\(\vec { j} \)+ lz\(\vec { k} \)) = \(\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ x & y & z \\ { p }_{ x } & { p }_{ y } & { p }_{ z } \end{matrix} \right| \)
 =(yp_z - zp_y)\(\vec { i} \) +(zp_x - xp_z)\(\vec { j} \)+(xp_y - yp_x)\(\vec { k} \)
From this relation, we conclude that 
lx= yp_z - zp_y, l_y= zp_x - xp_z and l_z = xp_y - yp_x
If the given particle moves only in the x - y plane, then z = 0 and p_z = 0 and hence,
\(\vec { l } \)=(xp_y-yp_x)\(\hat { k } \) ,which is only the .z-comonent of \(\vec { l } \)
It means that for a particle moving only in the x - y plane, the angular momentum has only the z-component

The tension in the string when it makes angle \(\theta\) with the vertical line is given by
\(T={mV^2\over r}+mgcos\theta\) (V is the instantaneous velocity)
\(T={0.15\times(2)^2\over 1}+0.15\times 10\times cos 48.2^o\)
=(0.6 + 1.5 x 0.67) N
= 1.6 N
Now let \(\overrightarrow{v_1}\) be the speed of the particle at the lowest point
\(\therefore V_1^2=V^2+2gr(1-cos \theta)\)
= (2)² + 2 x 10 x 1 (1 - cos 48.2⁰)
= 4 + 20(1 - 0.67)
= 4 + 20 x 0.33
= 4 + 6.6
= 10.6
\(\therefore V_1=\sqrt{10.6}=3.25ms^{-1}\)
The minimum value of V_I so that the particle is able to complete its vertical circle is  \(\sqrt{5gr}\)
\(\therefore \overrightarrow{v_{1}}_{min}=\sqrt{5\times10\times1}=7.07m/s\)
The value of \(\overrightarrow{v_1}\) obtained above is less than this minimum speed. The particle in the given case would not be able to complete its vertical circular path.

(i) Boys were courageous and have presence of mind.
(ii) Volume of water flowing per second,
\(V=20\quad L/min\)
\(\\ =\ \frac { 20\times 1000 }{ 60\times (100)^{ 3 } } =\frac { 1 }{ 3 } \times { 10 }^{ -3 }{ m }^{ 3 }/s\)
Radius of pipe, \(r=\frac { 4 }{ 2 } =2\quad cm=0.02m\)
Area of cross-section, \(a={ \pi r }^{ 2 }=\frac { 22 }{ 7 } \times (0.02)^{ 2 }m^{ 2 }\) 
Let v be the velocity of the flow of water, at given point. so volume , V = av
or     \(\frac { 1 }{ 3 } \times { 10 }^{ -3 }=\frac { 22 }{ 7 } =(0.02)^{ 2 }\times v\) 
\(\\ v=\frac { 7\times { 10 }^{ -3 } }{ 3\times 22\times (0.02)^{ 2 } } =0.2651m/s\)
\(\\ v=0.2651m/s\)
(iii) We should say no to crakers, because
(a) bursting of crackers pollutes the environment.
(b) crackers industry employs a lot of children fro the job and if we don't but cracker, they will be rescued.

Ajay is naughty but when he is made to realise his mistake, he is ready to apologise

Actual length of path travelled =23 km
Displacement = 10 km
Time taken =20 min =28/60 h
Average speed of taxi =\(\frac { actual \ path \ length }{ time \ taken } \)
= \(\frac { 23 }{ 28/60 } =49.3\quad km/h\)
Magnitude of average velocity = \(\frac { displacement }{ time } =\frac { 10 }{ 28/60 } =21.4\quad km/h\)

(iii) What will be the position-time curve for dropped packet?
In position-time curve, when a packet is dropped. Let us consider origin at the point where packet was dropped.

(i) Draw position-time graph for
(a) Accelerated motion  
(b) Retarded motion



(ii) Juggler throws the ball into the air with ball having initial velocity = u
At the highest point of its path velocity will be zero as at maximum height velocity =0
So final velocity=0
According to question juggler throws a ball each second so for "n" balls time taken to reach highest position (t)= 1/n
v=u+gt
0=u-g/n
-u=-g/n
u=g/n
S= ut-gt²/2
S=u/n-g/2n²
But u=g/n
S= g/n²-g/2n²
S=g/2n²

(a) Mean value of diameter,
 \({ D }_{ m }=\frac { 1.328+1.330+1.325+1.326+1.334+1.336 }{ 6 } \)
\(=\frac { 7.979 }{ 6 } =1.3298=1.330\)
(b) Taking D_m as the true value absolute errors in different observations are
 \({ \triangle }_{ 1 }=\left| 1.330-1.328 \right| =0.002\quad cm\)
\(\\ { \triangle }_{ 2 }=\left| 1.330-1.330 \right| =0.00\quad cm\)
\(\\ { \triangle }_{ 3 }=\left| 1.330-1.325 \right| =0.005\quad cm\)
\(\\ { \triangle }_{ 4 }=\left| 1.330-1.326 \right| =0.004\quad cm\)
\(\\ { \triangle }_{ 5 }=\left| 1.330-1.334 \right| =0.004\quad cm\)
\(\\ { \triangle }_{ 6 }=\left| 1.330-1.336 \right| =0.006\quad cm\)