Since, \(T=\propto \sqrt { l } \)
Therefore,
T = constant\(\sqrt{l}\)
\(\frac { { T }_{ f } }{ { T }_{ i } } =\sqrt { \frac { l+\frac { 44 }{ 100 } l }{ l } } =\sqrt { 1.44 } =1.2\)
Therefore, T_f = 1.2 T_i = T_i + 20% T_i

\(\vec { A } +\vec { B } =\left( 5\hat { i } +7\hat { j } -4\hat { k } \right) +\left( 6\hat { i } +3\hat { j } +2\hat { k } \right) =11\hat { i } +10\hat { j } -2\hat { k } \)
\(\vec { B } +\vec { A } =\left( 6\hat { i } +3\hat { j } +2\hat { k } \right) +\left( 5\hat { i } +7\hat { j } -4\hat { k } \right) =\left( 6+5 \right) \hat { i } +\left( 3+7 \right) \hat { j } +\left( 2-4 \right) \hat { k } =11\hat { i } +10\hat { j } -2\hat { k } \)
\(\vec { A } -\vec { B } =\left( 5\hat { i } +7\hat { j } -4\hat { k } \right) -\left( 6\hat { i } +3\hat { j } +2\hat { k } \right) =-\hat { i } +4\hat { j } -6\hat { k } \)
\(\vec { B } -\vec { A } =-\hat { i } +4\hat { j } -6\hat { k } \)
Note that the vector \(\vec A+\vec B\) and \(\vec B+\vec A \) are same and the vectors \(\vec A-\vec B\) and \(\vec B-\vec A \) are opposite to each other.

Let x₁ and x₂ be the elongation of springs from their equilibrium position (un-stretched position) due to the applied force F. Then, the net displacement of the mass point is
x = x₁ +x₂
Prom Hooke's law, the net force
\(F=-{ k }_{ S }\left( { x }_{ 1 }+{ x }_{ 2 } \right) \Rightarrow { x }_{ 1 }+{ x }_{ 2 }=-\cfrac { F }{ { k }_{ s } } \)

For springs in series connection
- k₁x₁ = - k₂r₂ = P
\(\Rightarrow { x }_{ 1 }=-\cfrac { F }{ { k }_{ 1 } } \) and \({ x }_{ 2 }=-\cfrac { F }{ { k }_{ 2 } } \)
Therefore, substituting equation (3) in equation (2), the effective spring constant can be calculated as
\(-\cfrac { F }{ { k }_{ 1 } } -\cfrac { F }{ { k }_{ 2 } } =\cfrac { F }{ k_{ s } } \)
\(\cfrac { 1 }{ { k }_{ s } } =\cfrac { 1 }{ { k }_{ 1 } } +\cfrac { 1 }{ { k }_{ 2 } } \) (or) \({ k }_{ s }=\cfrac { { k }_{ 1 }{ k }_{ 2 } }{ { k }_{ 1 }+{ k }_{ 2 } } \)
Suppose we have n springs connected in series, the effective spring constant in series is
\(\cfrac { 1 }{ { k }_{ s } } =\cfrac { 1 }{ { k }_{ 1 } } +\cfrac { 1 }{ { k }_{ 2 } } +\cfrac { 1 }{ { k }_{ 3 } } +...+\cfrac { 1 }{ { k }_{ n } } =\sum _{ i=1 }^{ n }{ \cfrac { 1 }{ { k }_{ i } } } \)
If all spring constants are identical i.e., k₁ = k₂ = ...= k_n = k then
\(\cfrac { 1 }{ { K }_{ s } } =\cfrac { n }{ k } \Rightarrow { k }_{ s }=\cfrac { k }{ n } \)
This means that the effective spring constant reduces by the factor n. Hence, for springs in series connection, the effective spring constant is lesser than the individual spring constants. From equation (3), we have,
k₁x₁ = k_2xx₂
Then the ratio of compressed distance or elongated distance x₁ and x₂ is
\(\cfrac { { x }_{ 2 } }{ { x }_{ 1 } } =\cfrac { { k }_{ 1 } }{ { k }_{ 2 } } \)
The elastic potential energy stored in first and second springs are \({ v }_{ 1 }=\cfrac { 1 }{ 2 } { k }_{ 1 }{ x }_{ 1 }^{ 2 }\) and \({ v }_{ 2 }=\cfrac { 1 }{ 2 } { k }_{ 2 }{ x }_{ 2 }^{ 2 }\) respectively. Then, their ratio is
\(\cfrac { { V }_{ 1 } }{ { V }_{ 2 } } =\cfrac { \frac { 1 }{ 2 } { k }_{ 1 }{ x }_{ 1 }^{ 2 } }{ \frac { 1 }{ 2 } { k }_{ 2 }{ x }_{ 2 }^{ 2 } } =\cfrac { { k }_{ 1 } }{ { k_{ 2 } } } \left( \cfrac { { x }_{ 1 } }{ x_{ 2 } } \right) ^{ 2 }=\cfrac { { k }_{ 2 } }{ k_{ 1 } } \)

Heat produced by the system, Q₁ = T₀S₀+\({1\over2}\)
T₀S₀+\({1\over2}\)T₀S₀
Q₂ = T_o (2S_o - S_o) = T_oS_o
Q₃ = 0
∴ Efficiency, \(η=1-{Q_2\over Q_1}\)
\(=1-{T_0S_0\over {3\over2}T_0S_0}\)
\(=1-{2T_0S̶_0\over 3T_0 S̶_0}⇒1-{2\over3}={3-2\over3}={1\over 3}\)
∴ Efficiency, η = \(1\over 3\) .

Time period T remains same, this is cos the period of oscillation of a liquid in a U-tube does not depend on the density of the liquid.

(i) The value of g varies with height has
g a\(\frac { 1 }{ { \left( R+h \right) }^{ 2 } } \ or\ g\ a\ \frac { 1 }{ { r }^{ 2 } } \)
Thus the graph of g versus V is the parabolic curve AB

(ii) The value of g varies with depth d as
\(g=g\left( 1-\frac { d }{ R } \right) \)i.e g\(\alpha \)(R-d)
Thus the graph of g versus depth d is the straight line AB.
 

Sound waves can be classified in three groups according to their range of frequencies:
(1) Infrasonic waves : Sound waves having frequencies below 20. Hz are called infrasonic waves. These waves are produced during earthquakes. Human beings cannot hear these frequencies. Snakes can hear these frequencies.
(2) Audible waves : Sound waves having frequencies between 20 Hz to 20,000 Hz (20kHz) are called audible waves. Human beings can hear these frequencies.
(3) Ultrasonic waves : Sound waves having frequencies greater than 20 kHz are known as ultrasonic waves. Human beings cannot hear these frequencies. Bats can produce and hear these frequencies.

Let V be the total volume of body
When body is floating in water, then
\(V {\rho_{body}} g =0.6\)
\(V \rho _{water }g \ \ (or) \rho _{water} = \frac{\rho _{body}}{0.6}\)
When body is floating in oil, then
\(V \rho_{body} g = 0.4 V \rho \ oil \ g\)
or \(\rho_{oil}=\frac{\rho_{body}}{0.4}\)
relative density of oil
\(= \frac{\rho_{oil}}{\rho _{water}}= \frac{\rho _{body }/0.4}{\rho_{body}/0.6}=\frac{6}{4}\)=1.5

(i) A tuning fork is struck on a rubber pad, the prongs of the tuning fork vibrate about their mean positions.
(ii) The prong vibrating about a mean position means moving outward and inward.
(iii) When a prong moves outward, it pushes the layer of air in its neighbourhood which means there is more accumulation of air molecules in this region.
(iv) Hence, the density and also the pressure increase. These regions are known as compressed regions or compressions.
(v) This compressed air layer moves forward and compresses the next neighbouring layer in a similar manner. Thus a wave of compression advances or passes through air.
(vi) When the prong moves inwards, the particles of the medium are moved to the right. In this region both density and pressure are low. It is known as a rarefaction or elongation.

When two or more waves. superimpose each other with slightly different frequencies, then a sound of periodically varying amplitude at a point is observed. This phenomenon is known as beats. The number of amplitude maxima per second is called beat frequency. If we have two sources, then their difference in frequency gives the beat frequency.
Number of beats per second
n = |f₁ - f₂| per second

Let the moment of inertia of the jester with his arms extended be I. As there is no external torque acting on the jester and the turn table, his total angular momentum is conserved. We can write the equation,
\(I_i \omega_i=I_f \omega_f\)
\(I \omega=\frac{1}{3} I\omega_f\)    \(\because (I_f=\frac{1}{3}I)\)
\(\omega_f=3\omega\)
The above result tells that the final angular velocity is three times that of initial angular velocity.

When the block slides, the force acting on the block is kinetic friction which is equal to f_k= μ_smg.
From Newton's second law ma = -μ_smg
The negative sign implies that force acts on the opposite direction of motion.
The acceleration of the block while sliding a =-μ_kg
The negative sign implies that the acceleration is in opposite direction of the velocity. Note that the acceleration depends only on g and the coefficient of kinetic friction μ_k. We can apply the following kinematic equation
v=u+at
The final velocity is zero
0=u-u_kgt
t=\(\frac { u }{ { \mu }_{ k }g } \).

Average velocity
\(\vec v_{avg}=\frac{\vec r_p-\vec r_O}{\Delta t}\)
Here \(\Delta t=5s\)
 \(\vec r_0-i ;\vec r_p=10\hat i\)
\(\vec v_{avg}=\frac{10\hat i}{5 sec}=2\hat icms^{-1}\)
The average velocity is in the positive x-direction.
The average speed total path length/time taken (the path is semi-circular)
= \(\frac{5\pi cm}{5s}=\pi cms^{-1}=3.14 cms^{-1}\)
Note that the average speed is greater than the magnitude of the average velocity.

Life of the man = 100 years
100 years includes 76 normal years and 24 leap years
Total no of days = 76\(\times\)365 + 24\(\times\)366 = 36524 days
Number of seconds = 36524\(\times\)24\(\times\)3600 = 3.155\(\times\)10° second
\(\text{Number of hearts beats}=\frac{Total\ no\ of\ seconds}{Time\ period\ of\ heart\ beat}=\frac{3.155\times10^9}{0.8s}=3.94\times10^9\)

Here M = 3 kg, R = 40 cm = 0.40 m, F = 30 N
Torque, \(\tau \)= F \(\times\) R = 30 \(\times\) 0.40 = 12 Nm
M.l. of the hollow cylinder about its own axis,
I = MR² = 3 \(\times\) (0.40)² = 0.48 kgm²
Angular acceleration, \(\alpha =\frac { \tau }{ I } \)= \(\frac {12}{0.48}\)= 25 rad s^-2
Linear acceleration, \(\alpha\) = R\(\alpha\) = 0.40\(\times\)25= 10 ms^-2

(i) The Law of conservation of linear momentum is a vector law. It implies that both the magnitude and direction of total linear momentum are constant. In some cases, this total momentum can also be zero.
(ii) To analyse the motion of a particle, we can either use Newton's second law or the law of conservation of linear momentum. Newton's second law requires us to specify the forces involved in the process. This is difficult to specify in real situations. But conservation of linear momentum does not require any force involved in the process. It is convenient and hence important.

Given: Speed of the bullet = 350 m/s
i.e., u = 350 m/s
s = 5 cm
v = 0 m/s
a = ?
Formula: v² = u² + 2as
\(\Rightarrow\) 0 = u² + 2as
(or)  u² = -2as (or) a = \(\frac{-u^2}{2s}\)
(or)  a = \(\frac{-350\times 350}{2\times 0.5}\) = -12.25\(\times\)10⁵ m/sec².

\({{1\ AU}\over{1\ parsec}}={{1.496\times{10}^{11}}\over{3.08\times {10}^{16}}}\)
\(={{1.496\times{10}^{11}\times{10}^{-16}}\over{3.08}}\)
\(={{1.496\times{10}^{-5}}\over{3.08}}\)
= 0.485\(\times\)10^-5
= 4.85\(\times\)10^-6 par sec.
4.85\(\times\)10^-6 parsec present in one astronomical unit.

(i) The energy produced by mechanical means is called mechanical energy.
(ii) It is classified into 2 types : (1) Kinetic energy (2) Potential energy.
(iii) The energy possessed by a body due to its motion is called kinetic energy. The energy possessed by the body by virtue of its position is called potential energy. SI unit of energy: N m (or) joule (J).