\(\frac { { T }_{ 2 }^{ 2 } }{ { T }_{ 1 }^{ 2 } } =\frac { { l }_{ 2 } }{ { l }_{ 1 } } =\frac { \left( \frac { l }{ 3 } \right) }{ l } =\frac { 1 }{ 3 } \quad or\quad \frac { { T }_{ 2 }^{ 2 } }{ { T }_{ 1 }^{ 2 } } =\frac { { (2) }^{ 2 } }{ 3 } or\ { T }_{ 2 }=\frac { 2 }{ \sqrt { 3 } } s\)

The moments of inertia will be different on account of different distances.

The angle is 180°, because force of friction always opposes the relative motion.

Distance between a compression and adjoining rarefaction is\(\frac {\lambda}{2}\)

For maximum work, \(\theta\) = 0° and for minimum work,  \(\theta\)= 90°.

The condition is that the lengths of the rods are inversely proportional to the coefficients of linear expansion of the materials of the rods

A straight line.

A gas has greatest internal energy and a solid has the least internal energy.

Time period of satellite is, given by \(T=2\pi\sqrt {\frac{(R+h)^3}{GM}}\)
Therefore, T will decrease if h decreases.

As force of attraction between molecules disappears, then the molecules will hit the wall with more speeds, hence , F = \(\frac{\Delta p}{\Delta t}\) , where F is average force on the wall due to the molecules.
\(\Delta\) p is change in momentum and \( \Delta \)t is the time duration . Due to increase in \(\Delta\) p, force F will also increase, hence pressure, p = \(\frac{F}{A}\) will increase. Here, A is area of one wall.

Yes, the dissipative forces become more important as the fluid velocity increases.
The viscous drag is given by
F =- \(\eta\) A  \(\frac{dv}{dx}\)
As the velocity of fluid increases, the velocity gradient increases and hence, viscous drag increases i.e. dissipative force also increases.

Kuldeep does not obey trafic rules.He must learn a lesson from this accident that rash driving is dangerous.

The relation is written by the boy
\(m=\frac { { m }_{ 0 } }{ (1-{ \nu }^{ 2 })^{ 1/2 } } \)
According to the principle of homogeneity of dimensions, the dimensions on either side of a relation must be same i.e., the powers of M, L, T on either side of a relation must be same.
Dimension of m is equal to the dimension of \( { m }_{ 0 } \), therefore, the denominator \({ (1-{ \nu }^{ 2 })^{ 1/2 } } \) should be dimensionless. In denominator 1 is dimensionless but factor \({ \nu }^{ 2 }\)is not dimensionless.
To make it dimensionless, we have to divide it by the same physical quantity with same power, therefore, it should be \({ { \nu }^{ 2 } }/{ { c }^{ 2 } }\), to become dimensionless.
Hence, the correct relation should be  \(m=\frac { { m }_{ 0 } }{ (1-\frac { { \nu }^{ 2 } }{ { c }^{ 2 } } )^{ 1/2 } } \)

The industrial revolution in England and Western Europe more than two centuries ago was triggered by some key scientific and technological advances.
A few of which are given below
(i) Discovery of electricity
(ii) Invention of powerloom, safety lamp, cotton gin, steam engine, etc.

Figure shows a tunnel dug along the diameter of the earth. Consider the case of a particle of mass m at a distance y from the centre of the earth. There will be a gravitational attraction of the earth on this particle due to the portion of matter contained in a sphere of radius y. The mass of the sphere of radius y is given by

M=Volume x density
or M=\(\frac { 4 }{ 3 } \pi { y }^{ 3 }\times d\)
(where d = density of earth).
This mass can be regarded as concentrated at the centre of the earth. The force F between this mass and the particle of mass m is given by
\(F=-\frac { GMm }{ { y }^{ 2 } } \)
Negative sign shows that the force is of attraction.
\(\therefore F=-G\left( \frac { 4 }{ 3 } \pi { y }^{ 3 }d \right) \frac { m }{ { y }^{ 2 } } =-G\times \left( \frac { 4 }{ 3 } \pi md \right) y\)
or F∝y.
The force is directly proportional to the displacement, hence the motion is simple harmonic motion.
Here ,the constant k=\(\frac { 4 }{ 3 } \pi mdG\)
The time period, T=\(2\pi \sqrt { (m/k) } \)
\(=2\pi \sqrt { \left( \frac { 3m }{ 4\pi mdG } \right) } =2\pi \sqrt { \left( \frac { 3 }{ 4\pi dG } \right) } \)
\(=\sqrt { \left( \frac { 3\pi }{ dG } \right) } =\sqrt { \left( \frac { 3\times 3.14 }{ 5.51\times { 10 }^{ 3 }\times 6.67\times { 10 }^{ -11 } } \right) } \)
= 42.2 minutes.

Consider velocity-time graph for uniform motion along a straight path. The graph is a straight line parallel to the time axis as shown in following Fig. Let A and B be two points on velocity-time graph corresponding to the instants t_l and t_2. As the motion is uniform, hence, AA_I = BB_I = V.
∴ Area under v-t graph between t₁ and t₂ = area ABB₁A₁
=AA₁ x A₁B₁ = V (t₂ - t₁)
But velocity is defined as v =\(\frac { Displacement }{ Time } =\frac { x_{ 2 }-x_{ 1 } }{ t_{ 2 }-t_{ 1 } } \)
∴ V (t₂ - t₁) = x₂ - x₁
∴ area ABB₁A₁ = (x₂ - x₁)
Hence, displacement of a particle in time interval (t₂ - t₁)is numerically equal to the area under velocity-time graph between the instants t₁ and t₂.

The measured (nominal) volume of the block is,
V = l x w x h
= (1.37 x 4.11 x 2.56) cm³
= 14.41 cm³
The least count of Vernier Caliper is ± 0.01 cm
\(\therefore\) Uncertain values can be written as
 l = (1.37 ± 0.01) cm
w = (4.11 ± 0.01) cm
h = (2.56 ± 0.01) cm
Lower limit of the volume of the block is,
 V_(min) = (1.37 - 0.01) x (4.11 - 0.01) x (2.56 - 0.01) cm³
= (1.36 x 4.10 x 2.55) cm³
= 14.22 cm³
This is 0.19 cm³ lower than the nominal measured value.
Similarly the upper limit can also be calculated as follows.
 V_(max) = (1.37 + 0.01) x (4.11 + 0.01) x (2.56 + 0.01) cm³
= (1.38 x 4.12 x 2.57) cm³
= 14.61 cm³
This is 0.20 cm³ higher than the measured value.
But we choose the higher of these two values as the uncertainty i.e. (14.41 ± 0.20) cm³.

We know that coefficient of real expansion of liquid (\(\gamma\)_r)= Coefficient of apparent expansion of the liquid (\(\gamma\)_a) + coefficient of volume expansion (\(\gamma\)_v).
i..e  \(\gamma +{ \gamma }_{ a }+{ \gamma }_{ v }={ \gamma }_{ a }+3a\)
Since the liquid is same in both the vessel, so value of Yr is same
For vessel A, \({ \gamma }_{ r }={ \gamma }_{ 1 }+{ \gamma }_{ a1 }=\gamma _{ 1 }+3a\)
For vesseal B = \({ \gamma }_{ r }={ \gamma }_{ 2 }+{ 3a }_{ 2 }\)
\(\therefore\)  \({ \gamma }_{ 1 }+3a={ \gamma }_{ 2 }+{ 3a }_{ 2 }\)
or \({ a }_{ 2 }=\frac { { \gamma }_{ 1 }-{ \gamma }_{ 2 } }{ 3 } +a\)

Numerical: Let I be the length of the wire at 20°C and ley the length at O°C. Then
l-l₀ = ∝l₀ ΔT = 20 ∝.l₀
Compressive strain =\(\frac { l-{ l }_{ 0 } }{ { l }_{ 0 } } \) =20∝ = 20 x 12 x 10^-6 = 2.4 x 10^-4
\(\Upsilon \)= \(\frac { Stress }{ Strain } \)
Stress = \(\Upsilon \) x strain = \(\frac{F}{A}\)
Hence, tension T=\(\Upsilon \)A x Strain
= 21 x 10¹¹ x 0.5 x 10^-2 x 2.4 x10^-4
= 2.52 x 10⁶ dyne
= 25.2 N
This is the tension in the wire when the temperature falls to O°C.

T₂ = 27°C = 27 + 273
= 300K
η = 40%, T₂₂ = ?
From \(η = 1-{T_2\over T_1}\)
\({T_2\over T_1}=1-η=1-{40\over 100}={60\over 100}={3\over 5}\)
\(T_1={5\over 3}T_2={5\over 3}\times300=500K\)
Increase in efficiency 10% of 40 = 4%
∴ New efficiency 'Y]' 40 + 4 = 44%
Let T₁ be the new temperature of the source
As \(η'=1-{T_2\over T_1'}\)
\({T_2\over T_1'}=1-η'=1-{44\over 100}={56\over 100}\)
\(T_1'={100\over 56}T_2={100\over 56}\times300=535.7K\)
∴ Increase in temp. of source
= 535.7 - 500 = 35.7 K

Let a be the angle of scattering of the coalesced mass (m + 2m) i.e. 3m and V be the velocity after the collision at the origin.

\(\therefore\) According to law of conservation of momentum, for x-component.
mv = 3mV cos \(\alpha\)
v = 3V cos \(\alpha\)                            (i)
For y-component
2mv = 3mV sin \(\alpha\)
2v = 3V sin \(\alpha\)                           (ii)
Dividing eq. (ii) by eq. (i) we get
\({ tan\quad \alpha =\frac { 2mv }{ mv } =2 }\)
\(\therefore \quad \alpha ={ tan }^{ -1 }(2)\)
= 63.4°
Squaring eq. (i) and eq. (ii) we get,
v² + (2v)² = (3V cos  \(\alpha\))² + (3V sin \(\alpha\))²
\(\Rightarrow\) v² + 4v² = 9V² cos² \(\alpha\) + 9 V² sin² \(\alpha\)
\(\Rightarrow\) 5v² = 9V²
\(\therefore\) \({ V }^{ 2 }=\frac { 5 }{ 9 } { v }^{ 2 }\)
or \(V=\frac { \sqrt { 5 } }{ 3 } v\)    (iii)
Now the K.E. before the collision = \(\frac { 1 }{ 2 } { mv }^{ 2 }+\frac { 1 }{ 2 } (2m){ v }^{ 2 }\)
\(=\frac { 3 }{ 2 } { mv }^{ 2 }\)
and the K.E. after the collision \(=\frac { 1 }{ 2 } (3m){ V }^{ 2 }\)
\(=\frac { 1 }{ 2 } \times 3m\times \frac { 5 }{ 9 } { v }^{ 2 }\)    [From (iii)]
\(=\frac { 5 }{ 6 } { mv }^{ 2 }\)
\(\therefore\) Loss of K.E. during the collission = \(\frac { 3 }{ 2 } { mv }^{ 2 }-\frac { 5 }{ 6 } { m }^{ 2 }\)
\(=\frac { 2 }{ 3 } { mv }^{ 2 }\)

The atmospheric pressure, P = 76 cm of mercury
(a) From figure (a),
Pressure head, h 20 cm of mercury
\(\therefore\) Absolute pressure p + h = 76 + 20 = 96 cm of mercury
Also, Gauge pressure h = 20 cm of mercury
From figure (b),
pressure head, h -18 cm of mercury
\(\therefore\) Absolutepressure p + h = 76 + (-18) = 58 cm of mercury
Also, Gauge pressure h = -18 cm of mercury
(b) When 13.6 cm of water is poured into the right limb of the manometer of figure (b), then, using the relation:
Pressure = pgh = p'g'h'
We get h' =  \(\frac{ph}{p'}=\frac{1\times 13.6}{13.6}=1\)cm of mercury
Therefore, pressure at the point B,
P_B = P + h' = 76 + 1 = 77 cm of mercury
If h" is the difference in the mercury levels in the two limbs, then taking PA = PB
\(\Rightarrow\) 58 + h" = 77 \(\Rightarrow\) h" = 77 - 58 = 19 cm of mercury.

Let m be the mass of the stick concentrated at C, the 50 cm mark, see fig.

For equilibrium about C', the 45 cm mark
10g (45-12) = mg (50-45)
10g 33 = mg\(\times\)5
⇒  m= \(\frac { 10\times 33 }{ 5 } \)
or  m = 66 grams

\(\frac { { A }_{ 1 } }{ { A }_{ 2 } } =\frac { 3 }{ 5 } \Rightarrow \sqrt { \frac { { I }_{ 1 } }{ { I }_{ 2 } } } =\frac { 3 }{ 5 }\)
Now, \(\frac { { I }_{ max } }{ { I }_{ min } } =\left( \frac { \sqrt { { I }_{ 1 } } +\sqrt { { I }_{ 2 } } }{ \sqrt { { I }_{ 1 } } -{ I }_{ 2 } } \right) =\left( \frac { \sqrt { { I }_{ 1 }/{ I }_{ 2 } } +1 }{ \sqrt { { I }_{ 1 } } /{ I }_{ 2 }-1 } \right) ^{ 2 }\)
\(\\ =\left( \frac { { 3 }/{ 5 }+1 }{ { 3 }/{ 5 }-1 } \right) ^{ 2 }=\frac { 64 }{ 4 } =\frac { 16 }{ 1 } \)

(a) As condition is, KE = 2PE
\(\frac { 1 }{ 2 } m{ \omega }^{ 2 }({ A }^{ 2 }-{ x }^{ 2 })=2\times \frac { 1 }{ 2 } m{ \omega }^{ 2 }{ A }^{ 2 }{ x }^{ 2 }\)
\(\\ \Rightarrow { A }^{ 2 }-{ x }^{ 2 }=2{ x }^{ 2 }\)
\(\Rightarrow x=\pm \frac { A }{ \sqrt { 3 } } \)
(b) This study shows that sum total of PE and KE of a particle in SHM stays constant at all positions and at all times. However, PE and KE both keep on changing with position or time. The same is true in day to day life. We can acquire one form of energy by spending some other form of energy and vice-versa.

Let
CV´= molar heat capacity of the first gas,
CV´´ = molar heat capacity of the second gas,
CV = molar heat capacity of the mixture,
and similar symbols for other quantities. Then,
γ= C′p/C′V=1.67
and
C′P=C′V+R
This gives
C′V=3/2R and C′P=5/2R
Similarly, γ = 1.4 gives C′′V=5/2R
and C′′P=7/2R
Suppose the temperature of the mixture is increased by dT. The increase in the internal energy of the first gas = n₁C′VdT.
The increase in internal energy of the second gas = n₂C′′VdT.
Thus, \(\left(n_1+n_2\right) C_V d T=n_1 C_V^{\prime} d T+n_2 C V " d T \)
or, \(C_V=\frac{n_1 C_v^{\prime}+n_2 C_v^{\prime \prime}}{n_1+n_2}\)
\( C_P=C_V+R=\frac{n_1 C_v^{\prime}+n_2 C_v^{\prime \prime}}{n_1+n_2}+R \)
\( =\frac{n_1\left(C_v^{\prime}+R\right)+n_2\left(C_v^{\prime \prime}+R\right)}{n_1+n_2} \)
\(=\frac{n_1 C_p^{\prime}+n_2 C_p^{\prime \prime}}{n_1+n_2} \ldots(2)\)
From (1) and (2),
\( Y=\frac{C_p}{C_V}=\frac{n_1 C_p^{\prime}+n_2 C_p^{\prime \prime}}{n_1 C_V^{\prime}+n_2 C_V^{\prime \prime}} \)
\( \frac{4 \times \frac{5}{2} R+2 \times \frac{7}{2} R}{4 \times \frac{3}{2} R+2 \times \frac{5}{2} R}=1.54 \)

Volume covered by the movement of smaller piston inwards is equal to volume moved out wards due to the larger piston.
\(\because \) water is incompressible
L₁A₁ = L₂A₂
\({ { L }_{ 2 } }=\frac { { A }_{ 1 } }{ { A }_{ 1 } } { L }_{ 1 }=\frac { \pi \left( \frac { 1 }{ 2 } \times { 10 }^{ -2 } \right) ^{ 2 } }{ \pi \left( \frac { 3 }{ 2 } \times { 10 }^{ -2 } \right) ^{ 2 } } \times 6\times { 10 }^{ -2 }\)
\(\\ { L }_{ 2 }=0.67\times { 10 }^{ -2 }m=0.67\ m\)
Atmospheric pressure is common to both piston and has been ignored.

\(\triangle l=12-10=2cm=0.02m,\ l=10cm=0.1m\) 
\(\\ a=1mm^{ 2 }=10^{ 6 }m^{ 2 } \ and \ Y=5\times 10^{ 8 }N/m^{ 2 }\)
\(\\ As, \ Y=\frac { F }{ a } \times \frac { l }{ \triangle l }\)
\( \\ \therefore F=\frac { Ya\triangle l }{ l } =\frac { 5\times 10^{ 8 }\times 10^{ -6 }\times 0.02 }{ 0.1 } =100N\)
KE of missile = elastic potential energy
Given m = 5g
From Eq (i), we have tension = 100N
\(\frac { 1 }{ 2 } mv^{ 2 }=\frac { 1 }{ 2 } \times tension\times extension\)
\(\\ v=\sqrt { \frac { tension\times 0.02 }{ m } } =\sqrt { \frac { 100\times 0.02 }{ 5/1000 } } =20m/s\)​​​​​​

From law of gravitation, \(F=\frac { GMm }{ { R }^{ 2 } } \)
Where, m = mass of object on surface of the earth.
R = radius of the earth.
M = mass of the earth.
Force experienced by object, F = mg
Using the two forces, we have
\(mg=\frac { GMm }{ { R }^{ 2 } } \quad or\quad M=\frac { gR^{ 2 } }{ G } \)

Here \(\mathbf{r}=\overline{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}\)
and \(\mathbf{F}=7 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}\)
We shall use the determinant rule to find the torque τ = r x F
\(\tau=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & -1 & 1 \\ 7 & 3 & -5 \end{array}\right|=(5-3) \hat{\mathbf{i}}-(-5-7) \hat{\mathbf{j}}+(3-(-7)) \hat{\mathbf{k}}\)
or \(\boldsymbol{\tau}=2 \overrightarrow{\mathbf{i}}+12 \hat{\mathbf{j}}+10 \hat{\mathbf{k}}\)

Given f_s = 980 N, R = Mg = 1200 kgf = 1200 x 9.8 N
Now, coefficient of static friction
i.e.  \({ \mu }_{ s }\) = \(\frac { { f }_{ s } }{ R } \)  = \(\frac { 980 }{ 1200\times 9.8 } \) 
= 0.83

Here, m = 1kg, u = 20 m/s
s = 10 cm = 0.1 m
v = 0, a= ? 
(a) As,
\(v^2=u^2+2 a s\)
\( \therefore 0=(20)^2+2 . a \cdot(0.1) \)
\( \Rightarrow a=-\frac{400}{2 \times 0.1}=-2000 \mathrm{~m} / \mathrm{s}^2 \)
b) v=u+a t
\( \therefore 0=20+(-2000) t \)
\( \Rightarrow t=\frac{20}{2000} \)
\(=\frac{1}{100}=0.01 \mathrm{sec}\)
c) Impulse = F. t
 =m(v-u)
 =1(0-20) 
 =-20 N.s.

m =10 kg, d = 0.4m, r = 0.2m
Revolutions per min, \(\nu =1000/min=\frac { 1000 }{ 60 } s,Linear\)
velocity, v=?,centripetal acceleration, a=?
\(\\ \omega =2\pi \nu =2\pi \times \frac { 1000 }{ 60 } =\frac { 100\pi }{ 3 } rad/s\)
\(\\ v=r\omega =0.2\times \frac { 100\pi }{ 3 } =\frac { 20\pi }{ 3 } m/s\)
\(\\ a=r\omega ^{ 2 }=0.2\times \left( \frac { 100\pi }{ 3 } \right) ^{ 2 }=\frac { 2000\pi ^{ 2 } }{ 9 } m/s^{ 2 }\)

(i) Mass vaccination is good as, it is used to make the socity free from the diseases like small pox.
(ii) Television for eradication of illiteracy and for mass communication of news and ideas is good as, it is a medium which is easily within the reach of common man and also they are very habitual to it.
(iii) Prenatal sex determination is bad because people are misusing it. Some of the people after determination of sex of child, think to abort. They do it especially with girl child.
(iv) Computer for increase in work efficiency is good as using the computer, a man can do much more work with greater efficiency and accuracy.
(v) Putting artificial satellite into orbits around the Earth is good for development as these satellites serve many purpose like remote sensing, weather foresting.
(vi) Development of nuclear weapons is bad as they can be used in mass destruction.
(vii) Development of new and powerful tool of chemical and biological warfare are bad, as they can also be used for mass destruction.
(viii) Purification of water for drinking purpose is good as we can save ourself from the diseases which we can have due to drinking of the water.
(ix) Plastic surgery is good as with the help of it a man or women can remove the skin defects occurring due to accident or some other reasons. It has some bad effects too but they are not very considerable.
(x) Cloning is good as far as animals are concerned. With the help of it, we can develop some species of animals which can be used to serve some specific purpose. But it is not good for human beings.

(i) Force on the floor by the pilot
= mg + ma
= m(g + a) = 60 (10 + 10)
= 1200 N (downward)
(ii) Force of helicopter on the surrounding air = (m₁ + m₂) (g + a)
= (500 + 60) (10 + 10)
= 11200N (downwards)
(iii) According to Newton's third law of motion, action and reaction are equal and opposite.
\(\therefore\) Force on the helicopter due to surrounding air = 11200 N (upwards).

For the body,
m₁ = 0.2 kg; \(\triangle\)T₁ = 100⁰ - 16⁰ = 84⁰C
S₁ = ?
For water,
m₂ = 0.5 kg; \(\triangle\)T₂ = 16⁰ - 10⁰ =6⁰C;
S₂ = 4.2 10³J/Kg/⁰C
From law of conservation of energy;
heat lost by body =  heat gained by water
i...e, m₁s₁ \(\triangle\)T₁ = m₂S₂ \(\triangle\)T₂
or S₁ = \(\frac { { m }_{ 2 }s_{ 2 }\triangle { T }_{ 2 } }{ { m }_{ 1 }\triangle { T }_{ 1 } } =\frac { 0.5\times \left( 4.2\times { 10 }^{ 3 } \right) \times 6 }{ 0.2\times 84 } \)
= 0.75 x 10³ J/Kg/⁰C

Graphs depicting variation of (i) gravitational potential energy (P.E.), (ii) kinetic energy (K.E.),and (iii) the total sum of potential and kinetic energies for a freely falling body are as shown in adjoining Fig. From the graphs, it is clear that:
(a) Gravitational potential energy decreases as the body falls downwards and is zero at the Earth.

(b) Kinetic energy increases as the body falls downwards and is maximum when the body just strikes the ground.
(c) The sum of kinetic and potential energies remains constant at all points during its free fall.

Here, \(\rho\) = 0.09 kg m^-3 
At S.T.P., Pressure P = 1.01 x 10⁵ pa.
According to kinetic theory of gases.
\(p=\frac { 1 }{ 3 } \rho { C }^{ 2 }\quad or\quad C=\sqrt { \frac { 3p }{ \rho } } \)
\(=\sqrt { \frac { 3\times 1.01\times { 10 }^{ 5 } }{ 0.09 } } =1837.5\quad { ms }^{ -1 }\)
Volume occupied by one mole of hydrogen at S.T.P. = 22.4 liters = 22.4 x 10^-3 m³
\(\therefore \) Mass of hydrogen, M = volume x density
= 22.4 x 10^-3 x 0.09
= 2.016 x 10^-3kg
Average K.E/mole =\(\frac { 1 }{ 2 } \)MC2
=\(\frac { 1 }{ 2 } \) x (2.016 x 10^-3) x (1837.5)²
= 3403.4 J

(a) False. The Young's modulus is defined as the ratio of stress to the strain within elastic limit. For a given stretching force elongation is more in rubber and quite less in steel. Hence, rubber is less elastic than steel.
(b) True. Stretching of a coil is determined by its shear modulus. When equal and opposite forces are applied at opposite ends of a coil, the distance as well as shape of helicals of the coil change and it involves shear modulus.

Progress in the field of science and technology is interrelated. Sometimes technology gives rise to new physics and at other times physics generates new technology. The discipline of thermodynamics arose mainly to understand and improve the working of heat engines. Similarly discovery of basic laws of electricity and magnetism led to development of wireless communication technology. Therefore, we can conclude that physics and technology are closely related.

Let us choose the nucleus of the hydrogen atom as the origin for measuring distance. Mass of hydrogen atom, m₁ = 1 unit (say) Since cholorine atom is 35.5 times as massive as hydrogen atom,
mass of cholorine atom, m₂ = 35.5 units
Now, x₁ = 0 and x₂ = 1.27\(\mathring { A } \) =1.27\(\times\)10^-10 m
Distance of centre of m~ss of HCl molecule from the origin is given by
X =  \(\frac { { m }_{ 1 }{ x }_{ 1 }+{ m }_{ 2 }{ x }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } =\frac { 1\times 0+35.5\times 1.27\times 10 }{ 1+35.5 } m\)
= \(\frac { 35.5\times 1.27 }{ 36.5 } \times { 10 }^{ -10 }m\) 
= 1.235\(\times\)10-10m
= 1.235\(\mathring { A } \)

Let v₁,v₂ be the velocities of S wave and P wave and t₁, t₂ be the time taken by these waves to reach the seismograph
l= distance of occurrence of earthquake from the seismograph
\(v_{ 1 }t_{ 1 }=v_{ 2 }t_{ 2 }\)
\( \Rightarrow v_{ 1 }=4kms^{ -1 },v_{ 2 }=8kms^{ -1 }\)
\( \Rightarrow 4t_{ 1 }=8t_{ 2 }\Rightarrow t_{ 1 }=2t_{ 2 }\)
\( t_{ 1 }-t_{ 2 }=4min=\ 240s\)
\(On\ solving\ Eqs.\ (i)\ and\ (ii),\ t_{ 2 }=240s\)
\(\Rightarrow t_{ 1 }=2t_{ 2 }=2\times 240=480s\)
\(\Rightarrow l=v_{ 1 }t_{ 1 }=4\times 480=1920km\) 

(i) For a simple pendulum k is proportional to m the mass of the particle hence \(\frac { m }{ k } \) becomes constant and does not affect the time period.
(ii) If we replace \(sin\theta \approx \theta \) for large angles, then actually \(sin\theta <\theta \)
Now since this factor is multiplied to the restoring force mg\(sin\theta \) is replaced by \(mg\theta \) which means an effective reduction in g for large angles. hence, there is an increase in time period T over that given by the formula              
 \(T=2\pi \sqrt { \frac { l }{ g } } \)
as compared to the case which it is assumed \(sin\theta \simeq \theta \)               
(iii) Yes, since the motion of hands of a wristwatch to indicate time depends on action of the spring and has nothing to do with acceleration due to gravity.
(iv) In a free fall the effective g = 0,i.e. gravity disappears 
Time period \(T=2\pi \sqrt { \frac { l }{ g } } =2\pi \sqrt { \frac { l }{ 0 } } =\infty \)
Frequency, \(v=\frac { 1 }{ T } =0\)
i.e. frequency of oscillation is zero.

Given, Displacement s = 30 m
Force, F = 30 kg - wt = 30 x 10 = 300N
Angle between force and ground ,  \(\theta\) = 60^o
The work done by the gardener,
W = F.s = Fscos \(\theta\)  = 300 x 30 x cos 60^o
W = 4500 J

(a) When two bodies at different temperatures T₁ and T₂ are brought in thermal contact, heat flows from the body at the higher temperature to the body at the lower temperature till equilibrium is achieved, i.e., the temperatures of both the bodies become equal. The equilibrium temperature is equal to the mean temperature (T₁ + T₂)/2 only when the thermal capacities of both the bodies are equal.
(b) The coolant in a chemical or nuclear plant should have a high specific heat. This is because higher the specific heat of the coolant, higher is its heat-absorbing capacity and vice versa. Hence, a liquid having a high specific heat is the best coolant to be used in a nuclear or chemical plant. This would prevent different parts of the plant from getting too hot.
(c) When a car is in motion, the air temperature inside the car increases because of the motion of the air molecules. According to Charles’ law, the temperature is directly proportional to pressure. Hence, if the temperature inside a tyre increases, then the air pressure in it will also increase.
(d) A harbour town has a more temperate climate (i.e., without the extremes of heat or cold) than a town located in a desert at the same latitude. This is because the relative humidity in a harbour town is more than it is in a desert town.

Let the speed of the flow be v.
Given, diameter of tap = d = 1.25 cm
Volume of water flowing out per second.
Q = v \( \times\) \(\frac{\pi d^2}{4} \) \(\Rightarrow\) v = \(\frac{4Q}{d^2\pi}\)
Estimate Reynold's number, Re = \(\frac{4 \rho Q}{\pi d \eta}\)
Q = 0.48 L / min 
= 8 \(\times\) 10^-3 L/s
= 8 \(\times\)10^-6 m³/s
Re =  \(\frac{4\times10^3\times8\times10^{-6}}{3.14\times1.25\times10^{-2}\times10^{-3}} \)
Re = 815                        [ i.e below 1000, the flow is steady ]
After some time, when
Q = 3L/min 
= 5\(\times\)10^-5m³/s,
Re =  \(\frac{4\times10^3\times5\times10^{-5}}{3.14\times1.25\times10^{-2}\times10{-3}} \)
= 5095
ஃ The flow will be turbulent.

\( T \propto r^{3 / 2} \)
\( \therefore \frac{T_1}{T_2}=\left(\frac{10^{13}}{10^{12}}\right)^{3 / 2}=10 \sqrt{10} . \)

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.

Given, initial velocity (u) = 40m/s
Height of the hall (H) = 25m
Let the angle of projection of the ball be \(\theta \), when maximum height attained by it be 25m.
Maximum height attained by the ball
\(H=\frac { { u }^{ 2 }{ sin }^{ 2 }\theta }{ 2g } \Rightarrow 25=\frac { { (40) }^{ 2 }{ sin }^{ 2 }\theta }{ 2\times 9.8 }\)
\(or\ { \sin }^{ 2 }\theta =\frac { 25\times 2\times 9.8 }{ 1600 } =0.3068\)
\(or\ \sin\theta =0.5534=\sin{ 33.6 }^{ 0 }\)
\(or\ \theta ={ 33.6 }^{ 0 }\)
\( \therefore \text{ Horizontal range (R)}=\frac { { u }^{ 2 }\sin2\theta }{ g }\)
\( \\ =\frac { { (40) }^{ 2 }sin2\times { 33.6 }^{ 0 } }{ 9.8 } =\frac { 1600\times sin{ 67.2 }^{ 0 } }{ 9.8 } \)
\(\\ =\frac { 1600\times 0.9219 }{ 9.8 } =150.5m\)

(i) \(4.6\times 0.128\) = 0.5888 = 0.59.
The result has been rounded off to have two significant digits (as in 4.6).
(ii) \(\frac { 0.9995\times 1.53 }{ 1.592 } \) = 0.96057 = 0.961
The result has been rounded off to three significant digits (as in 1.53).
(iii) 876 + 0.4382.
876 + 0.4382 = 876.4382 = 876
As, there is no decimal point in 876, therefore, result of addition has been rounded off to no decimal point.