A vector cannot be zero if one of its components is non-zero.

Radius of gyration of a given body depends upon the choice of rotation axis. However, for a given axis of rotation, the radius of gyration of a body has a fixed value.

Since, a =\({F\over m}\) On doubling the force, the acceleration will also be doubled.

The phase difference between two given particles = \(\frac{2\pi}{\lambda} \times \frac{\lambda}{2} = \pi\) radian, i.e., the two particles are in mutually opposite phase conditions. 

Percentage increase in speed \(=\frac{v_e-v_0}{v_0}\times100=(\frac{v_e}{v_0}-1)\times100\)
\(=(\sqrt 2-1)\times100=41.4\%\)

C.V. Raman (the great Indian scientist) received Nobel prize for his work on molecular spectra.

We have , 0.25 x 6 x 10²³ molecules, each of volume
Molecular volume = 2.5 x 10^-7 m³
Supposing, ideal gas law is valid.
Final volume =  \(\frac { { V }_{ in } }{ 100 }\) = \(\frac { { (3) }^{ 3 }\times { 10 }^{ -6 } }{ 100 }\)
\(\approx\) 2.7\(\times\)  10 ^-7 m³ 
Which is about the molecular volume. hence, intermolecular forces cannot be neglected. Therefore, the ideal gas situation does not hold.

When a gas in cylinder is compressed by pushing in a piston, the speed of the molecules or their kinetic energy increases. This increases the temperature of the gas.

Bulk modulus of rubber (b) = 9.8 x 10⁸ N/m²
Density of seawater (p) = 10³ kg/m³
Percentage decrease in volume
\(\left( \frac { \Delta V }{ V } \times 100 \right) =0.1\ or\ \frac { \Delta V }{ V } =\frac { 0.1 }{ 100 } \)
\(\\ or \frac { \Delta V }{ V } =\frac { 1 }{ 1000 } \)
Let the rubber ball be taken up to depth h.
Change in pressure (p) = hpg
Bulk modulus \((B)=\frac { p }{ (\Delta V/V) } =\frac { hpg }{ (\Delta V/V) } \)
\(or\ h=\frac { B\times (\Delta V/V) }{ pg } =\frac { 9.8\times 10^{ 8 }\times \frac { 1 }{ 1000 } }{ 10^{ 3 }\times 9.8 } =100m\) 

We can calculate the spring constant of spring, as it ios extended by 20 cm under 60 N force.
F = kx \(\Rightarrow\)60 = k\(\times\) 20 \(\times\)10^-2 
k = 300 N/m
At a force of 20 N, the extension in spring is 
F = kx \(\Rightarrow\) 20 = 300x
x = \(\frac { 2 }{ 30 } =\frac { 1 }{ 15 } m\)

We can see that the object P covers a distance of 10 m in every 15 min. In other words, it covers equal distance in equal intervals of time. So, the motion of object P is uniform.
On the other hand, the object Q covers 7 m from 9:30 am to 9:45 am, 4 m from 9:45 am to 10:00 am and so on. In other words, it covers unequal distances in equal intervals of time. So, the motion of object Q is non-uniform.

Strain and specific gravity.

Line of sight is defined as an imaginary line joining an object and an observer’s eye. When we observe nearby stationary objects such as trees, houses, etc. while sitting in a moving train, they appear to move rapidly in the opposite direction because the line-of-sight changes very rapidly.
On the other hand, distant objects such as trees, stars, etc. appear stationary because of the large distance. As a result, the line of sight does not change its direction rapidly.

Let the body be projected with velocity u inclined at angle \(\theta\) with the horizontal. The horizontal and vertical components of velocity and acceleration are
u_x, a_x and u_y, a_y
where u_x = u cos \(\theta\) u_y = u sin\(\theta\) ,a_x = 0,a_Y = -g
g is the acceleration due to gravity.

The coordinates of O are (0, 0) considering horizontal motion.
The position of the body after time t has coordinate (x, y);
where  x (t)=x_o + u_x t + \({1\over2}a_x t^2\)
Substituting for various factors
x (t)=x_o + U cos \(\theta\) . t + \({1\over2}\) x 0 x t²
or x (t) =u cos \(\theta\) . t
\(t={x(t)\over u \ cos \theta}\)            ............(i)
Considering the vertical motion
y (t) Y (0) + u_y +\({1\over2}a_y t^2\)
or  y (t)=0+ U sin \(\theta\) . t - \({1\over2}\) gt²
or  y (t)=U sin \(\theta\) . t - \({1\over2}\) gt²  ............(ii)
Substituting for t from equation (i) in equation (ii), we get
\(y(t)=u \ sin \theta ({x(t)\over u \cos \theta})-{1\over 2}g({x(t)\over u \ cos \theta})^2\)
\(\Rightarrow \ y(t)=x(t)tan \theta -{1\over2}g{x^2(t)\over u^2 \ cos^2 \theta}\) ...............(iii)
This is an equation of parabola. Thus, the path of a projectile is a parabola.
Maximum height attained. At the maximum height, the vertical component of velocity becomes zero. Now using the equation of motion.
\(h={v^2_y-u^2_y\over 2a_y}\)
We have maximum height
\(\therefore h_{max}={0^2-(u \ sin \theta)^2\over 2(-g)}\)
or \(H={u^2sin^2\theta \over 2g}\)  .....................(iv)
Time for maximum height. Using equation of motion v = u + at
or v_x = u_x + a_y t
we have 0 =u sin \(\theta\) - gt
or t=\(u \ sin \theta \over \ g\) .....................(v)
Horizontal range. Let the horizontal range be x. Since there is no acceleration in the horizontal direction so
x = x (0) + u_x t +  \({1\over2}a_x t^2\)
As x (0) = 0, u_x = U cos \(\theta\), a_x= 0 and it is the total time of the flight which is twice the time for maximum height because body takes same time in rising to and falling from the highest point.
Hence,   t = \({2u \ sin \theta \over g}\)
\(\therefore \) x = o + u cos \(\theta\) . t = u cos\(\theta\)\(({2u \ sin \theta \over g})\)
or x \(={u^2\over \ g}(2 \ sin \theta cos \theta)\)
\(\Rightarrow \ x={u^2\over g}sin \ 2 \theta\) ....................(vi)
Maximum horizontal range. From equation (vi) for x to be maximum, the value of sin 2\(\theta\) should be maximum which is 1,
hence         \(x_{max}={u^2\over g}\) ....................(vii)
For this x_max' sin 2\(\theta\)= 1\(\Rightarrow\) \(\theta\) = 45°
Therefore, the horizontal range will be maximum if the angle of projection is 45° or \(\pi\over 4\) radians.
Time of flight of the projectiles. The projectile after completing its flight returns back to the same horizontal level from which it was projected. Therefore, the vertical displacement in the whole flight is zero. Considering vertical motion.
y(t) = y(0)  +u_yt + \({1\over2}a_yt^2\)
Now y(t) = 0,y(o) = 0,u_y = u sin \(\theta\) , a_y = -g
Then 0 = 0 + u  sin \(\theta\) . T-\({1\over2}gT^2\)
\(\Rightarrow T({u \ sin \theta -{1\over2}gT})=0\)
Therefore T = 0
and  U sin \(\theta -{1\over 2}gT=0\Rightarrow u\ sin \theta ={1\over2}gT\)
or gT = 2u sin \(\theta\)
\(T={2u \ sin \theta \over g}\) ...........(viii)
Equation (viii) gives the total time of flight. This is twice the time for maximum height.

Let A be the top of the building at a height X above the ground. Be is the window of height 2m. Let t_l be the time taken by the ball to fail from A to B. Then the time taken to fall through a distance 
\((x + 2) m is (t_1 + 0.2) s\). Then
x = \(\frac{1}{2}\times 10\times t_1^2 \) ...(i)
and \(x+2=\frac{1}{2}\times 10\times(t_1+0.2)^2 \) ...(ii)
From Eqns. (1) and (2), we have
\(5t_1^2+2=5t_1^2+0.2+2t_1\)
or \(1.8=2t_1\)
∴ t₁=0.9s
∴ \( t_1=0.9s\)
∴ x = \(\frac{1}{2}\times 10\times (0.9)^2=4.05m\)
Let V₁ be the speed acquired by the ball as it reaches the bottom C of the window.
Then
\( v_1=0+10 x 1.1=11ms^{-1}\)
During the downward journey the ball experiences an acceleration equal to g and strikes the ground with speed V. Since it undergoes a perfectly elastic collision its speed reverses.
The time taken in falling from C to ground to back is 2 s. Therefore, time taken to fall from C to ground must be 1 s. Hence
\( X -(X+2)=11 \times 1 +\frac{1}{2} \times 10(1)^2\)
or \(X -6.05=11+5=16\)
∴ X = 22.05 m
The height of the building is 22.05 m and the bottom of the window is at height of 16 m [= (22.05 - 4.05) m] from the ground.

Equilibrium: A body or a system of particles is said to be in a state of equilibrium if inspite of a number of forces-or torques acting on it, the body or the system of particles remains in its original state of rest or of uniform motion (translational or rotational or both). Thus, equilibrium state means that acceleration (both linear as well as angular) of the body/system must be zero. Hence, equilibrium is of two types:
(i) Equilibrium of rest: If a given system remains in a state of rest and does not change its position inspite of number of forces acting on it, it is said to be in an equilibrium of rest e.g., our house, our school etc.
(ii) Equilibrium of motion: If a given system maintains its state of uniform motion, translational or rotational or combined, under the action of a number of forces then it is said to be in an "equilibrium of motion" e.g., our Earth, the planetary system, electrons revolving around the nucleus of an atom. For a state of equilibrium of motion, the value of linear momentum and/ or angular momentum of the system should have a finite and constant value.
Conditions for complete equilibrium: For complete equilibrium condition for translational equilibrium and condition for rotational equilibrium both must be fulfilled. The conditions are
(a) For translational motion we know that \(\frac { d\vec { p } }{ dt } =\sum { \vec { F_{ ext } } } \)
For equilibrium \(\vec { p } \) = a constant or \(\frac { d\vec { p } }{ dt } \)=0 or \(\sum { \vec { F_{ ext } } } =0\)
Hence for translational equilibrium, the vector sum of all the external forces acting on the system/ body under discussion must be zero
(b) For rotational motion, we that \(\frac { d\vec { L } }{ dt } =\sum { \vec { \tau _{ ext } } } \)
For equilibrium \(\vec { L} \) = a constant or \(\frac { d\vec { L } }{ dt } =0\quad or\quad \sum { \vec { { \tau } } _{ ext } } =0\)
Hence for rotational equilibrium, the vector sum of all the external torques acting on the system/body must be zero

Here,
T₁ = 200°C = 473 K and T₂ = 100°C = 373 K
∴ Efficiency of engine \(η={W\over Q_1}=\left(T_1-T_2\over T_1\right)={473-373\over 473}=0.21\)
W = 0.21 Q1 = 21% of Q
Thus, engine will convert 21% of heat used for doing work.

(a) Let a drop of a liquid L be poured on a solid surface S placed in air A. If T_SL, T_LA and T_SA be the surface tensions corresponding to solid-liquid layer, liquid-air layer and solid-air layer respectively and \(\theta\) be the angle of contact between the liquid and solid, then
       
\(T_{LA}\cos\theta+T_{SL}=T_{SA}\)
\(\Rightarrow \cos\theta=\frac{T_{SA}-T_{SL}}{T_{LA}}\)
For the mercury-glass interface, T_SA < T_SL, Therefore, cos \(\theta\) is negative. Thus \(\theta\) is an obtuse angle. For the water-glass interface, T_SA > T_SL, Therefore cos \(\theta\) is positive. Thus, \(\theta\) is an acute angle.
(b) Water on a clean glass surface tends to spread out i.e., water wets glass because force of cohesion of water is much less than the force of adhesion due to glass. In case of mercury force of cohesion due to mercury molecules is quite strong as compared to adhesion force due to glass. Consequently, mercury does not wet glass and tends to form drops.
(c) Surface tension of liquid is the force acting per unit length on a line drawn tangentially to the liquid surface at rest. Since t; .is force is independent of the area of liquid surface therefore, surface tension is also independent of the area of the liquid surface.
(d) We know that the clothes have narrow pores or spaces which act as capillaries. Also, we know that the rise of liquid in a capillary tube is directly proportional to cos \(\theta\) (Here \(\theta\) is the angle of contact). As \(\theta\) is small for detergent, therefore cos \(\theta\) will be large. Due to this, the detergent will penetrate more in the narrow pores of the clothes.
(e) We know that any system tends to remain in a state of minimum energy. In the absence of any external force for a given volume of liquid its surface area and consequently. Surface energy is least for a spherical shape. It is due to this reason that a liquid drop, in the absence of an external force is spherical in shape.

If  \({ v }_{ 1 },{ v }_{ 1 }{ ,T }_{ 1 }and{ v }_{ 2 },{ v }_{ 2 }{ ,T }_{ 2 }\) are the frequencies, velocity and tension in the first and second strings, respectively, then
\(\frac { { v }_{ 2 } }{ { v }_{ 1 } } =\frac { { v }_{ 2 }/2L }{ { v }_{ 1 }/2L } =\frac { { v }_{ 2 } }{ { v }_{ 1 } } \Rightarrow \quad \frac { { v }_{ 2 } }{ { v }_{ 1 } } =\frac { \sqrt { { T }_{ 2 }/\mu } }{ \sqrt { { T }_{ 1 }/\mu } } \)
Given that the tension in one string is 2.0% larger than the other
\({ T }_{ 2 }={ T }_{ 1 }+\frac { 2 }{ 100 } { T }_{ 1 }=1.02{ T }_{ 1 }\)
The ratio of frequencies
\(\frac { { v }_{ 2 } }{ { v }_{ 1 } } =\sqrt { \frac { 1.02{ T }_{ 1 } }{ { T }_{ 1 } } } =1.01\quad Hz\)
Now, solve for the frequency of the tightened string.
\({ v }_{ 2 }=1.01{ v }_{ 1 }=1.01\times 420=424.2\quad Hz\)
Thus, the beat frequency,
\({ v }_{ beat }={ v }_{ 2 }-{ v }_{ 1 }=424.2-420=4.2\quad Hz\) 

\(v_{1}=\frac{v}{4 \times 15}, v_{2}=\frac{v}{2 \times 30.5}=\frac{v}{61} \)
\(m=v_{1}-v_{2}=\frac{v}{60}-\frac{v}{61}=v \times \frac{1}{61 \times 60} \)
\(5=\frac{v}{61 \times 60}\)
or = v = 5 x 16 x 60 cm/s
\(\therefore \quad v_{1} =\frac{v}{60}=\frac{5 \times 61 \times 60}{60}=305 \mathrm{~Hz}\)
\(v_{2} =\frac{v}{61}=\frac{5 \times 60 \times 61}{61}=300 \mathrm{~Hz} \)

(i) \(\sin \omega t+\cos \omega t\) is a periodic function, it can also be written as \(\sqrt{2} \sin (\omega t+\pi / 4)\).
\( \text { Now } \sqrt{2} \sin (\omega t+\pi / 4)=\sqrt{2} \sin (\omega t+\pi / 4+2 \pi) \)
\(=  \sqrt{2} \sin [\omega(t+2 \pi / \omega)+\pi / 4]\)
The periodic time of the function is \(2 \pi / \omega\).
(ii) This is an example of a periodic motion. It can be noted that each term represents a periodic function with a different angular frequency. Since period is the least interval of time after which a function repeats its value, sin ωt has a period T₀ = 2π/ω ; cos 2 ωt has a period π/ω =T₀ /2; and sin 4 ωt has a period 2π/4ω = T₀ /4. The period of the first term is a multiple of the periods of the last two terms. Therefore, the smallest interval of time after which the sum of the three terms repeats is T₀ , and thus, the sum is a periodic function with a period 2π/ω.
(iii) The function e –ωt is not periodic, it decreases monotonically with increasing time and tends to zero as t → ∞ and thus, never repeats its value.
(iv) The function log(ωt) increases monotonically with time t. It, therefore, never repeats its value and is a nonperiodic function. It may be noted that as t → ∞, log(ωt) diverges to ∞. It, therefore, cannot represent any kind of physical displacement.

\( V=250 \mathrm{cc}=250 \times 10^{-6} \mathrm{~m}^3 \)
\( \mathrm{P}=10^{-3 \mathrm{~mm}}=10^{-3} \times 10^{-3} \mathrm{~m} \)
\( =\left(10^{-6} \times 13600 \times 10\right) \)
\(=136 \times 10^{-3} \text { Pascal } \)
\( \mathrm{T}=27^0 \mathrm{C}=300 \mathrm{k} \)
\( \mathrm{n}=\frac{\mathrm{PV}}{R T} \)
\( =\frac{136 \times 10^{-3} \times 250 \times 10^{-6}}{8.3 \times 300}=1.36 \times 10^{-8}\)
No. of molecules
\(=1.36 \times 10^{-8} \times 6 \times 10^{23}  \)
\(=8.17 \times 10^{15}\)

(i) -148^o F
(ii) 32^oZ

Total cross-sectional area of the femurs is A = 2 × 10 cm² = 20 × 10^–4 m² . The force acting on them is F = 40 kg wt = 400 N (taking g = 10 m s^–2). This force is acting vertically down and hence, normally on the femurs. Thus, the average pressure is
\({ P }_{ av }=\frac { F }{ A } =\frac { 400 }{ 20\times { 10 }^{ -4 } } \)

Given, mass(m) = 14 .5kg
Length of wire (l ) = 1 m
Angular frequency (v) = 2 revls
Angular velocity (\(\omega \)) = 2\(\pi \)v
= 2\(\pi \)\(\times \)2 rad/s = 4\(\pi \) rad/s
 
Area of cross-section of wire (A) = 0.065 cm²
= 6.5 \(\times \)10^-6 m²
Young's modulus for steel (Y) = 2 \(\times \) 10¹¹ N/m².
At lowest point of the vertical circle, T - mg = ml\({ \omega }^{ 2 }\)
or  T= mg +  m\({ \omega }^{ 2 }\)
= (14.5\(\times \)9.8)+14.5\(\times \)1\(\times \)\(​​{ (4\pi ) }^{ 2 }\)
= 14.5(9.8+16\({ \pi }^{ 2 }\))
= 14.5(9.8\(\times \)16\(\times \)9.87)          [\(\because \) \({ \pi }^{ 2 }\)=9.87]
= 14.5\(\times \) 167.72N=2431.94 N
Young's modulus (Y) =\(\frac { Stress }{ Strain } =\frac { (T/A) }{ \Delta l/l } =\frac { Tl }{ A.\Delta l } \)
\(\therefore \Delta l=\frac { T.l }{ A.Y } =\frac { 2431.94\times 1 }{ 6.5\times { 10 }^{ -6 }\times 2\times { 10 }^{ 11 } } \)
= 1.87\(\times \)10^-3 m =1.87 mm

Young's modulus \((Y)=\frac { Stress }{ Longitudinal\ strain } \)
For same longitudinal strain \(Y\alpha stress\)
\(\frac { { Y }_{ stress } }{ { Y }_{ rubber } } =\frac { { (stress) }_{ steel } }{ { (stress) }_{ rubber } } \)
But          \({ Y }_{ steel }>{ Y }_{ rubber }\)
\(\frac { { Y }_{ steel } }{ { Y }_{ rubber } } >1\)
Therefore from Eq(i) we get
\(\frac { { (stress) }_{ steel } }{ { (stress) }_{ rubber } } >1\)
\(or { (stress) }_{ steel }>(stress)_{ rubber }\)

To solve this question, we use the Kepler’s third law. This states that the square of the period of the revolution of all the planets about the sun is directly proportional to the cube of the mean distance between the planets and the sun. This can be mathematically given as
T2 ∝ R3
⇒T2=kR3 where  T is the period of the revolution of a planet around the sun,  R is the mean radius of the planet to the sun.
Hence, we can write by comparison between two planets, that
\(\frac{T_1^2}{T_2^2}=\frac{R_1^3}{R_2^3}\)
We can use any planet with a known distance and period as the second planet. We choose earth.
\(\frac{T_N^2}{T_E^2}=\frac{R_N^3}{R_E^3}\) where the subscript N and E stands for Neptune and Earth respectively.
For earth, the period is 1 year. Hence, write that
\( \frac{165^2}{1^2}=\frac{R_N^3}{R_E^3} \)
\( \Rightarrow R_N^3=165^2 R_E^3\)
Hence, by finding the cube root of both sides, we have
\(R_N=\sqrt[3]{165^2} R_E\)
⇒RN=30RE
Radius of the earth is about 1.50×10¹¹m
Hence, RN=30(1.50×1011)=4.5×10¹²m

If r is the distance of the given point from the centre of the earth, then gravitational potential at the point.
\(V=-\frac { GM }{ r } =-5.12\times { 10 }^{ 7 }\ J/kg\)
Acceleration due to gravity at this point,
\(g=\frac { GM }{ { r }^{ 2 } } =6.4\ m/{ s }^{ 2 }\)
\(\\ Clearly,\ \frac { \left| V \right| }{ g } =\frac { GM/r }{ GM/{ r }^{ 2 } } =r\)
\(\\ Thus,\ r=\frac { 5.12\times { 10 }^{ 7 }J/kg }{ 6.4m/{ s }^{ 2 } } =8\times { 10 }^{ 6 }m=8000\ km\)
Obviously, height of the point from the earth's surface
= (r-R) = 8000km - 6400km = 1600 km.

The nuclear decay may be represented as follows
Fe⁵⁷\(\rightarrow\) Fe⁵⁷ + hv (\(\gamma \)-ray photon)
According to de-Broglie hypothesis, momentum of a photon of energy y E is 
\(p=\frac { E }{ c } =\frac { 14.4\times 1.6\times { 10 }^{ -16 }J }{ 3\times { 1 }0^{ 8 }{ ms }^{ -1 } } \)
\(\\ p=\ 7.68\ \times { 10 }^{ -24 }kg { ms }^{ -1 }\)
By conservation of momentum, the momentum of daughter nucleus, p=momentum of \(\gamma \) -ray photon
\(=d\ 7.68 \ \times { 10 }^{ -24 }kg \ { ms }^{ -1 }\)
The recoil energy of the nucleus will be
\(k= \ \frac { { p }^{ 2 } }{ 2m } =\frac { ({ 7.68 \ \times { 10 }^{ -24 }) }^{ 2 } }{ 2\times \ 56.935 \ \times 1.66 \times { 10 }^{ -27 } }\)
\( \\ =0.32\times { 10 }^{ -21 }\)
\(\\ =\frac { 0.312\times { 10 }^{ -21 } }{ 1.6\times { 10 }^{ -16 } } keV\)
\(K=1.95{ \times 10 }^{ -16 }keV\)

Force F = ma, therefore force acting on a particle in unaccelerated (a = 0) motion is zero.
As car is moving with a constant velocity, therefore, its acceleration is zero. i.e., a = 0 therefore, net force acting on the car F = ma = 0.

This is an example of uniform circular motion. Here R = 12 cm. The angular speed ω is given by
The angular speed \(\omega\) is given as
\(\omega ={ 2\pi }/{ T }={ 2\pi \times 7 }/{ 100 }=0.44rad/s\) and linear speed \(\nu \)is\(\nu =\omega R=0.44\times 12=5.3{ cm\ s }^{ -1 }\)
The direction of velocity \(\nu \)is along the tangent to the circle at every point. The acceleration is directed towards the centre of the circle. Since, this direction changes continuously acceleration, here is not a constant vector.
However, the magnitude of acceleration is constant.
a = ω² R = (0.44 s^–1) 2 (12 cm) = 2.3 cm s^-2 

Let x₁,x₂ and x₃ be distances covered in three parts of the motion
For the first of the motion, we have
u = 0,t = 10s,v = 20m/s
As  v = u + at
 \(\therefore \ 20=0+a1\times 0\)
Acceleration a=2m/s²
Distance,   \({ x }_{ 1 }=ut+\frac { 1 }{ 2 } { at }^{ 2 }\)
\( =0\times 10+\frac { 1 }{ 2 } \times 2\times (10)^{ 2 }=100\quad m\)
For second Part of the motion, we have 
\({ x }_{ 1 }=100m,{ x }_{ 3 }=40\quad m\)
AS   \({ x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 }=640\ \)
\(\therefore \quad 100+{ x }_{ 2 }+40=640\)
\({ x }_{ 2 }=500m\)
The distance is covered with a uniform speed of 20m/s 
 \(\\ \therefore \)  Time taken =\(\frac { 500 }{ 20 } =25s\)
For third part of the motion, we have
u = 20m/s ,v = 0,x = x₃= 40m
As    \({ v }^{ 2 }-{ u }^{ 2 }=2ax\)
\(\\ { (0) }^{ 2 }-(20)^{ 2 }=2\times a\times 40\)
or   \(a=\frac { 400 }{ 80 } =-5m/s^{ 2 }\)
\(\therefore \)   Retardation = 5m/s²
Time taken,   \(t=\frac { v-a }{ a } =\frac { 0-20 }{ -5 } =4s\)
Total Time taken = 10 + 25 + 4 = 39s

\(Fractional\quad error=\frac { \triangle { D }_{ mean } }{ D } =\pm \frac { 0.004 }{ 1.330 } =\pm 0.003\).

(i) Velocity at any instant t is given by v = Aω cos ωt
Here A = 10cm,ω = \(\frac{2\pi}{T}=\frac{2\pi}{10}\)
When t = 2s, v = 10\(\times\)\(\frac{2\pi}{10}cos(\frac{2\pi}{10}\times2)\)
= 2\(\pi\) cos (0.4 \(\pi\))
= 1.942 cm/s.
(ii) Acceleration at any instant t is given by
a = - Aω² sin ωt
= -10\((\frac{2 \pi}{10})^2sin(0.4\pi)\)
= - 3.755 cm/ s²
acceleration is numerically equal to 3.754 cm/ s² and is directed towards the mean position.

Let V be the volume of the load attached to lower end of steel wire of Young's modulus \(\Upsilon \), area of cross-section A and length of wire L. Let \(\rho \)  be the relative density of rod.
Then, \(\Upsilon =\frac { FL }{ Ax_{ 1 } } =\frac { V\rho gL }{ Ax_{ 1 } } \)     ..........(i)
When load is immersed in liquid, then
\(\Upsilon =\frac { (V\rho g-V\times 1\times g) }{ Ax_{ 2 } } \)   .....(ii)
From (i) and (ii),
\(\frac { \rho }{ { x }_{ 1 } } =\frac { \rho -1 }{ { x }_{ 2 } } \)  or  \(\rho \)x₂ = \(\rho \)x₁ - x₁
or x₁ = \(\rho \)(x₁-x₂) or \(\quad \rho =\frac { { x }_{ 1 } }{ { x }_{ 1 }-{ x }_{ 2 } } \).

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

(a) B
(b) C
(c) Slower
(d) Yes, in the interval 2 to 3 s

Escape velocity, \(v=\sqrt {2gR}=\sqrt {\frac{2GM}{R}}.\)
Thus escape velocity of a planet depends upon (i) its mass (M) and (ii) its size (R). As different planets have different masses and sizes, so they have different escape velocities.

Four fundamental forces present in nature are:
(i) Gravitational force
(ii) Electromagnetic force
(iii) Weak nuclear force
(iv) Strong nuclear force.

2 beats

(i) Given, T = 27 \(^0\)C
= ( 273.15 + 27 )
= 300.15K
Average thermal energy , E =  \(\frac{3}{2}\)k_BT
( where, kB = Boltzman constant
= 1.38\(\times\)10^-23 JK^-1 )
E = \(\frac{3}{2}\)\(\times\)1.38\(\times\)10^-23\(\times\)300.15
= 6.21 \(\times\)10^-21 J
(ii) At the temperatures , T = 107 K
Average thermal energy , E = \( \frac{3}{2}\)k_BT
= \( \frac{3}{2}\)\(\times\)1.38\(\times\)10^-23\(\times\)6000
= 1.241\(\times\)10^-19 J
(iii) At temperature , T = 10⁷K
Average thermal energy,
E = \(\frac{3}{2}\)k_BT
= \(\frac{3}{2}\)\(\times\)1.38\(\times\)10^-23\(\times\)10⁷
= 2.07\(\times\)10^-16 J

Given, work done (W) = - 22.3 J
Work done is taken negative as work is done on the system.
In an adiabatic change, \(\Delta \)Q = 0
Using first law of thermodynamics
\(\Delta \).U = \(\Delta \)Q - W = 0 -(- 22.3) = 22.3J
For another process between states A and B,
Heat absorbed (\(\Delta \)Q) = + 9.35 cal
= + (9.35 x 4.19) J = + 39.18 J
Change in internal energy between two states via different paths are equal.
\(\because \)   \(\Delta \)U = 22.3 J
\(\therefore \)   From first law of thermodynamics,
\(\Delta \)U = \(\Delta \)Q - W
or W = \(\Delta \)Q -\(\Delta \)U
= 39.18 - 22.3
= 16.88J \(\approx \)16.9J

(a) The pressure of liquid column is given by \(p=h\rho g\), where h is depth, \(\rho \)i s density and g is acceleration due to gravity.
Therefore, pressure of liquid column increases with depth. The height of blood column in human body is more at feet than at the brain. Therefore, the blood pressure in humans is greater at the feet than the brain.
(b) The density of air is maximum near the surface of the earth and decreases rapidly with height. At a height of 6 km, the density of air decreases to nearly half its value at the seal level. Beyond 6km height, the density of air decreases very slowly with height.Hence, the atmospheric pressure at a height of about 6km decreases to nearly half of its value at the sea level.
(c) When force is applied on a liquid, the pressure is transmitted equally in all directions inside the liquid. Therefore, hydrostatic pressure has no fixed direction and hence, it is a scalar quantity.

Rolling motion, i.e. (Rotation + Translation)

When a man is standing on a weighing scale, it will read the normal reaction R as apparent weight.
Given , mass of man (m) = 70 kg
In each case the weighing scale will read the reaction R, i.e. the apparent weight.
(iv) Acceleration of the lift when it is falling freely under gravity
a = g (\(\downarrow\) )
\(\therefore\)  Normal reaction, R = m ( g - g ) = 0
\(\therefore\)  Reading on weighing scale = 0

zero, 3.6 N