Young's modulus of a material decreases with rise in temperature.

Because they have zero terminal velocity.

3 beats/s

x = -b and  x = -2b

Inertia and elasticity are the properties which are required by a system to oscillate.

(i) Here F = 10 N,\(\triangle l=0.1m,m=4kg\) 
\(k=\frac { F }{ \triangle l } =\frac { 10 }{ 0.1 } =100Nm^{ -1 }\)
(ii) Here F = 10 N,\(\triangle l=0.1m,m=4kg\)
\(y=\frac { mg }{ k } =\frac { 4\times 10 }{ 100 } =0.4m\)
(iii) Here F = 10 N,△l = 0.1m,m = 4kg
\(T=2\pi \sqrt { \frac { m }{ k } } =2\times \frac { 22 }{ 7 } \sqrt { \frac { 4 }{ 100 } } =1.26s\)
(iv) Here F = 10 N,△l = 0.1m,m = 4kg
Frequency, \(v=\frac { 1 }{ T } =\frac { 1 }{ 1.26 } =0.8\quad Hz\)          

For an adiabatic process, \(\triangle Q=0\) .So, if a process is carried very fast so that heat cannot transferred from system to surroundings and vice-versa, it is an adiabatic process.

No, the process of transfer of heat by convection is based on the fact that a liquid becomes lighter .on becoming hot and rise up. In condition of weightlessness, this is not possible. So, transfer of heat by convection is not possible in the earth satellite.

If the volumes are same, then height of the liquid will be highest in which the cross-sectional area is least at the top. So, the vessel having least cross-sectional area at the top possess more pressure at the base(∵p=ρgh)

Force is given by
\(F=\frac { GM_{ e }m }{ R^{ 2 } } =\frac { GM_{ p }m }{ ({ { D }_{ p } }/{ 2 })^{ 2 } } =\frac { 4GM_{ p }m }{ { D }_{ p }^{ 2 } } \)
\(\frac { F }{ m } =\frac { 4GM_{ p } }{ { D }_{ p }^{ 2 } } \)

Yes,as v = \(\sqrt {\frac{GM}{r}}\) ,v depends only upon r.For a particular orbit,r is constant and so is v.

Here, I = 50 kg-m², ω= 5rps  = 5 × 2 π rad/s
Angular momentum, L = Iω = 50 × 10π = 500 πJ−s

Given :P20 kg ms \({ }^{-1}, m=0.1 \mathrm{~kg}\)
Kinetic energy \(K=\frac{1}{2} m v^2\) and
Momentum p=m v or \(v=\frac{p}{m}\)
Eliminating v,
\(K=\frac{1}{2} m \times\left(\frac{p}{m}\right)^2=\frac{p^2}{2 m}\)
\(\therefore K=\frac{(20)^2}{2 \times 0.1}=2000 J=2 \times 10^3 J\)

Porcelain objects are wrapped in paper or straw before packing to reduce  the chances during transportation. During transportation  sudden jerks or even fall can take place. Forces are created at the point of collision and the force takes longer time to reach the porcelain objects through paper or straw for same change in momentum as  F = \({ \Delta p }/{ \Delta t }\) and therefore a lesser force acts on object.

Horizontal range,
  R = \(\frac { { u }^{ 2 }sin2\theta }{ g } or\quad 3=\frac { { u }^{ 2 }sin6{ 0 }^{ o } }{ g } =\frac { { u }^{ 2 } }{ g } \sqrt { 3/2 } \)
or   \(\frac { { u }^{ 2 } }{ g } 2\sqrt { 3 } \)
Since, the muzzle velocity is fixed
Therefore, maximum horizontal range,
  \({ R }_{ max }=\frac { { u }^{ 2 } }{ g } 2\sqrt { 3 } =3.464km\)
So, the bullet cannot hit the target.

 Both these statements are not true , because
(i) Its magnitude  can be zero.
(ii) Its magnitude is either less than or equal to the distance travelled by the object.

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.

Solid angle, relative density, strain, Reynold's number and Poisson's ratio.

We know that,\(v=\omega \sqrt { \left( { a }^{ 2 }-{ y }^{ 2 } \right) } \)
Further \(\omega =\frac { 2\pi }{ T } \)
\(v=\frac { 2\pi }{ T } \sqrt { ({ a }^{ 2 }-{ y }^{ 2 }) } =\frac { 2\times 3.14 }{ \left( \frac { 11 }{ 7 } \right) } \sqrt { [({ 1.0) }^{ 2 }-({ 0.6) }^{ 2 })] } \)
= 3.2 m/sec.
Kinetic energy at this displacement is given by
\(K=\frac { 1 }{ 2 } { mv }^{ 2 }\)
\(=\frac { 1 }{ 2 } \times 0.8\times { (3.2) }^{ 2 }=4.1joule\) 

Here, Young's modulus, \(\Upsilon \) = 2 x 10¹¹Nm^-2
Area of cross-section, A = 5 x 10^-5 m²
Let the initial length of wire be L. Then, increase in length of wire,
ΔL= L
Now, \(\Upsilon \)=\(\frac { F\times L }{ A\times \triangle L } \)
∴ F = \(\frac { \Upsilon \times A\times \triangle L }{ L } \)
⇒ F = \(\frac { 2\times 10^{ 11 }\times 5\times 10^{ -5 }\times L }{ L } \)
or F = 10⁷ N.

The density \(\rho\) of a wire of mass M, length L and diameter 'd' is given by
\(\rho = \frac{4M}{\pi d^{2}L}= \frac{4m}{\pi d^{2}}\)
Now \(\rho_{A} = \sqrt{ \frac{T}{m_{A}}}\)
and \(v_{B}=\sqrt{ \frac{T}{m_{B}}}\)
∴ \( \frac{v_{A}}{v_{B}}=\sqrt{ \frac{m_{B}}{m_{A}}}= \frac{d_{B}}{d_{A}}\)
but \(v_{A} = v\lambda_{A}\) and \(v_{A}=v\lambda_{B}\) n being the frequency of the source.
Hence \(\frac{\lambda_{A}}{\lambda_{B}}=\frac{v_{A}}{v_{B}}=\frac{d_{A}}{d_{B}}=\frac{0.5 \times 10^{-3}}{10^{-3}}=0.5\)

When a body floats in a liquid with a part submerged in the liquid, the weight of the liquid displaced by the submerged part is always equal to the weight of the body.
Let   V = volume of the body
 \(\sigma\) = density of its material
 \(\rho\) = density of the liquid in which the body floats such that its
volume V' is outside the liquid.
Then volume of the body inside the liquid = V - V'
Weight of the displaced liquid = (V - V') Pg
Also weight of the body = V \(\sigma\) g
For the body to float,
weight of the liquid displaced by the submerged part = weight of the body
i.e., (V - V') \(\rho g\) = V \(\sigma\) g
or      \(V'=\frac{(\rho-\sigma)V}{\rho}\) 

\(\lambda \)_m1 = 1.56 um, T₁ = 2000 K
\(\lambda \)_m2 = 1.8 um, T₂ = ?
\({ \lambda }_{ m }=\frac { b }{ T } \)
\(\therefore\)  \(\frac { { \lambda }_{ m1 } }{ { \lambda }_{ m2 } } =\frac { { T }_{ 2 } }{ { T }_{ 1 } } \)
\(\Rightarrow\) \(\frac { 1.56 }{ 1.8 } =\frac { { T }_{ 2 } }{ 2000 } \)
\(\Rightarrow\) \({ T }_{ 2 }=\frac { 1.56\times 2000 }{ 1.8 } \)
\(\therefore\)  \({ T }_{ 2 }=1733.3K\)

(i) 1 : 1
(ii) 4 : 1

\(\frac { g }{ 16 } \)

Let the acceleration of the centre of mass of disc be a, then , Ma = F - f 
The angular acceleration of the disc is a =a/ R (if there is no sliding).
Then, \(\left( \frac { 1 }{ 2 } M{ R }^{ 2 } \right) \alpha \) = Rf \(\Rightarrow \) Ma =2f
Thus, f = F /3.Since, there is no sliding.
\(\Rightarrow \) f <\(\mu \) F\(\le \) 3\(\mu \) Mg

(a) The maximum height is given by
\(h_m=\frac { \left( { \nu }_{ 0 }^{ 2 }\sin2\theta _{ 0 } \right)^2 }{ 2g } =\frac { ( 28\times \sin60^{ \circ })^2 }{ 2(9.8) } =m\)
\(= \frac{14 \times 14}{2 \times 9.8} = 10.0 m\)
(b) The time taken to return to the same level is Tf = (2 v_o sin θ_o )/g = (2 × 28 × sin 30° )/9.8 = 28/9.8 s = 2.9 s
(c) The distance from the thrower to the point where the ball returns to the same level is 
\(R=\frac { \left( { \nu }_{ 0 }^{ 2 }\sin2\theta _{ 0 } \right) }{ g } =\frac { 28\times 28\times \sin60^{ \circ } }{ 9.8 } =69m\)

Since, \({ e }^{ -\left( az/\theta \right) }\) is dimensionless, we have \(aZ/\theta =1\)  
or     \(a=\frac { \theta }{ Z } =\frac { K }{ L } =\left[ { L }^{ -1 }K \right] \)
We find that a/b = dimensions of p and b =\(\left[ { ML }^{ -1 }{ T }^{ -2 } \right] \)
Therefore, dimensional formula of p is obtained as
 \(p=\frac { a }{ \left[ { ML }^{ -1 }{ T }^{ -2 } \right] } =\frac { \left[ { L }^{ -1 }K \right] }{ \left[ { ML }^{ -1 }{ T }^{ -2 } \right] } =\left[ { M }^{ -1 }{ L }^{ 0 }{ T }^{ 2 }K \right] \) 

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 \({ a }_{ 2 }=\frac { { \gamma }_{ 1 }-{ \gamma }_{ 2 } }{ 3 } +a\)

Frequency of SONAR, \(v=40kHz=40\times { 10 }^{ 3 }Hz\)
Speed of observer/enemy's submarine
\(360km/h=\frac { 360\times 1000 }{ 60\times 60 } =100\quad m/s\)
Given, speed of sound wave in water =1450m/s
As observer is moving towards stationary source,hence, apparent frequency observed will be
\({ v }^{ ' }=\left( \frac { v+{ v }_{ 0 } }{ v } \right) v=\frac { (1450+100) }{ 1450 } \times 40\times { 10 }^{ 3 }\)
The wave is reflected by the submarine.
Now, submarine of theenemy willa ct as a source and SONAR will be observer.
Hence, apparent frequency observed
\({ v }^{ ,, }=\frac { v\times v }{ v-{ v }_{ s } } =\frac { 1450\times 4.276\times 1{ 0 }^{ 4 } }{ 1450-100 } \)
\(=4.59\times { 10 }^{ 4 }Hz=45.9kHz\) 

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

Mass of the block, m =15kg
Coefficient of friction between the block and the trolley
\(\mu =0.18\)
Acceleration of the trolley, a = 0.5m/s²
Time, t = 20 s
(i) As block is placed on the trolley, therefore, friction force is applied on the block by the trolley
F = ma = \(15\times 0.5=7.5N\)
Force on the block is the friction applied by trolley on the block, its direction is in the direction of motion of the trolley.
For the stationary observer on the ground, the block will appear to move with acceleration initially and then with uniform velocity as given in the question.
(ii) For an observer on the trolley, the block is always at rest either initially or finally. Trolley always becomes intertial frame with respect to block because both have the same acceleration initially and same velocity finally.

Let the frequency of vibration of the string be given by 
\(\nu =Kl^{ a }m^{ b }T^{ c }\) ......(i)
where K = a dimensionless constant
Dimensions of the various quantities are
\(\nu =\left[ T^{ -1 } \right] ,l=\left[ L \right] ,T=\left[ T \right] \)
Force =\(\left[ MLT^{ -2 } \right] \)
and \(m=\frac { mass }{ length } =\left[ ML^{ -1 } \right] \)
Substituting these dimensions in equation(i), we get
\(\left[ T^{ -1 } \right] ={ \left[ L \right] }^{ a }\left[ ML^{ -1 } \right] ^{ b }\left[ MLT^{ -2 } \right] ^{ c }\)
or \(\left[ M^{ 0 }L^{ 0 }T^{ -1 } \right] =\left[ { M }^{ b+c }{ L }^{ a-b+c }{ T }^{ -2c } \right] \)
Equating the dimensions of M,L and T, we get
b + c = 0, a - b + c=0 and - 2c = - 1
on solving,\(a=-1,b=-\frac { 1 }{ 2 } and\quad c=\frac { 1 }{ 2 } \)
\(\therefore \quad \left( \nu \right) =Kl^{ -1 }m^{ { -1 }/{ 2 } }T^{ { 1 }/{ 2 } }\quad or\quad \left( \nu \right) =\frac { K }{ l } \sqrt { \frac { T }{ m } } \)