We know that C_p - C_v = R
C_p - C_v = 5000
Dividing by C_v we get
\({C_P\over C_V}-1={5000\over C_V}\)
\(1.6-1={5000\over C_V}\)
\(0.6={5000\over C_V}\)
\(∴\ C_V={5000\over 0.6}=8333.33 J\ kg^{-1}k^{-1}\)
Now C_p - C_v = 5000
C_p - 8333.33 = 5000
C_P = 8333.33 + 5000
= 13333.33 J kg^-1 k^-1

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

Mass of child, M = 30kg = 30 x 10³g
Fall in temperature,
\(\Delta T=101-98=3^{ 0 }F\)
\(\\ =3\times \frac { 5 }{ 9 } =\frac { 5 }{ 3^{ 0 } } C\)
Specific heat of human body, 
c= specific heat of water
= 1 cal g^-1 0C^-1
Heat lost by child in the form of evaporation of sweat,
\(Q=Mc\Delta T=30\times 10^{ 3 }\times 1\times \frac { 5 }{ 3 } \)
= 50000 cal
If M' gram of sweat evaporates from the body of the child, then heat gained by sweat
Q = M'L= M' x 580 Cal
Heat gained = Heat lost
M' x 580 = 50000
\(\Rightarrow \quad M'=\frac { 50000 }{ 580 } =86.2g\) 
Time taken by sweat to evaporate = 20min
Rate of evaporation of sweat = \(\frac { 86.2 }{ 20 } \)
= 4.31g min^-1

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.

44 N

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 = \(\frac { displacement }{ time } =\frac { 10 }{ 28/60 } =21.4\quad km/h\)

(a) Here, height(h)=1 km = 1000m, g = 10m/s²
Velocity attained by the rain drop in freely falling through a height h.
\(v=\sqrt { 2gh } =\sqrt { 2\times 10\times 1000 }\)
\( \\ =100\sqrt { 2 } m/s\)
\(\\ =100\sqrt { 2 } \times \frac { 60\times 60 }{ 1000 } Km/h\)
\(\\ =360\sqrt { 2 } km/h\approx 510km/h\) 
(b) Diameter of the drop(d) = 2r = 4mm
Radius of the drop(r) = 2mm = 2 x 10^-3m
Mass of a rain drop(m)=Vx\(\rho \)
\(=\frac { 4 }{ 3 } \pi { r }^{ 3 }\rho =\frac { 4 }{ 3 } \times \frac { 22 }{ 7 } \times (2\times { 10 }^{ -3 })^{ 3 }\times { 10 }^{ 3 }\)
\( \approx 3.4\times { 10 }^{ -5 }kg\)
\(\\ Momentum \ of \ the \ rain \ drop(\rho )=mv \)
\(\\ =3.4\times { 10 }^{ -5 }\times 100\sqrt { 2 }\)
\(\\ \approx 4.7\times { 10 }^{ -3 }{ kg }^{ -m/s }\)
(c) Time required to flatten the drop
= time taken by the drop to travel the distance equal to the diameter of the drop near the ground
\(t=\frac { d }{ v } =\frac { 4\times { 10 }^{ -3 } }{ 100\sqrt { 2 } } =0.028\times { 10 }^{ -3 }s\)
\(\\ =2.8\times { 10 }^{ -5 }s\)
(d) Force exerted by a rain drop
\(F=\frac { Change \ in \ momentum }{ Time } \)
\(=\frac { \rho -0 }{ t } =\frac { 4.7\times { 10 }^{ -3 } }{ 2.8\times { 10 }^{ -5 } } \approx 168N\)
(e) Radius of the umbrella(R) = \(\frac { 1 }{ 2 } m\)
\(\therefore \ Area \ of \ the \ umbrella(A)=\pi { R }^{ 2 }\)
\(\\ =\frac { 22 }{ 7 } \times { (\frac { 1 }{ 2 } ) }^{ 2 }=\frac { 22 }{ 28 } =\frac { 11 }{ 14 } \approx 0.8{ m }^{ 2 }\)
Number of drops striking the umbrella simultaneously with average of 5cm or 5 x 10^-2m
\(=\frac { 0.8 }{ { (5\times { 10 }^{ -2 }) }^{ 2 } } =320\)
Net force exerted on umbrella 
= 320 x 168 = 53760 N

Given, Radius of olive oil, r = 0.25 mm = 0.025 cm
Radius of circular film, R = 10 cm and Molecular size, t = ?
We know that,
\(t=\frac { Volume\ of\ oil\ drop }{ Area\ of\ film } =\frac { \frac { 4 }{ 3 } \pi { r }^{ 3 } }{ \pi { R }^{ 2 } }\)
\( \\ t=\frac { \frac { 4 }{ 3 } \times \pi \times { 0.025 }^{ 3 } }{ \pi { \left( 10 \right) }^{ 2 } }\)
\( \\ =\frac { 4 }{ 3 } \times { \left( 25 \right) }^{ 3 }\times { 10 }^{ -11 }\)
\(\\ =2.08\times { 10 }^{ -7 }cm\)
Thus, the molecular size of olive oil is \(2.08\times { 10 }^{ -7 }cm\)

A vector can be non-zero even if one of its components is zero.

The time period is increased by a factor of \(\sqrt2\)

Node: It is a point on stationary wave at which amplitude of vibration of the particle is zero.
Antinode: It is a point on stationary wave at which amplitude of vibration of the particle is maximum.

Since \(k=\frac { { p }^{ 2 } }{ 2m }\ or\ p=\sqrt { 2mk } \)
As k is same for both bodies, \(p\infty \sqrt { m } \), i.e, heavier body has more momentum than the lighter body.

The ratio of r.m.s velocities is given as
\(\frac { { { C }_{ 1 } } }{ { C }_{ 2 } } =\sqrt { \frac { { M }_{ 2 } }{ { M }_{ 1 } } } =\sqrt { \frac { { \rho }_{ 2 } }{ { \rho }_{ 1 } } } \)
\(\frac { { C }_{ 1 } }{ { C }_{ 2 } } =\sqrt { \frac { 9 }{ 6 } } =\sqrt { 3 } :\sqrt { 2 } \)

The resultant amplitude will be
y=\(\sqrt{y_2^1+y_2^2}=\sqrt{9+16}=\sqrt{25}=\)5cm

π rad,i.e.180^o

Brass is a good conductor of heat. It quickly conducts away the heat. So, the paper does not alter its ignition point easily. On the other hand, wood is a bad conductor of heat and is unable to conduct away the heat. So, the paper quickly reaches its ignition point and is charred.

Yes, it is perfect correct to call heat as the energy in transit because it is continuously flowing on account of temperature different between bodies or parts of a system.

Two streamlines can never cross each other because if they cross them at the point of intersection, there will be two possible direction of flow of fluid which is impossible for streamlines.

When mass M is placed on the piston, the excess pressure, p=Mg/A. As the pressure is equally applicable from all the direction on the sphere, hence there will be decrease in volume due to decrease in raius sphere. Volume of the sphere, \(V=\frac { 3 }{ 4 } \pi { R }^{ 3 }\)
Differentiating it , we get,
\(\Delta V=\frac { 4 }{ 3 } \pi (3{ R }^{ 2 })\Delta R=4\pi { R }^{ 2 }\Delta R\)
\(\\ \frac { \Delta V }{ V } =\frac { 4\pi { R }^{ 2 }\Delta R }{ \frac { 4 }{ 3 } \pi { R }^{ 3 } } =\frac { 3\Delta R }{ R } \)
We know that, \(B=\frac { P }{ dV/V } =\frac { Mg }{ A } \diagup \frac { 3\Delta R }{ R } \)
\(or\ \frac { \Delta R }{ R } =\frac { Mg }{ 3BA } \) 

9.789\({ m }/s^{ 2 }\)

Given: The mass of each ball is 0.05 kg and the magnitude of speed of each ball is 6 ms^-1 .
Initial momentum of each ball is given as,
p i =mv
Where, m is the mass of each ball and v is the velocity of each ball.
By substituting the given values in the above expression, we get
p i =mv =0.05×6 =0.3  kgms^-1
After collision, the balls change their directions of motion without changing the magnitudes of their velocity.
Final momentum of each ball is given as,
p f =− p i
By substituting the given values in the above expression, we get
p f =−0.3  kgms^-1
Impulse imparted to each ball is given as,
I=Δp = p f − p i
By substituting the given values in the above expression, we get
I=−0.3−0.3 =−0.6  kgms^-1
The negative sign indicates that the impulses imparted to the balls are opposite in direction.
Thus, the impulse imparted to each ball is −0.6  kgms^-1.

1.496 x 10¹¹m

Displacement from mean position.
\(=A-\frac { A }{ 2 } =\frac { A }{ 2 } \)
Now y=A cos ωt
\(or\quad \frac { A }{ 2 } =Acos\frac { 2\pi }{ T } t\)
\(or\quad cos\frac { 2\pi }{ T } t=\frac { 1 }{ 2 } \)
\(\Rightarrow cos\frac { 2\pi }{ T } t=cos\frac { \pi }{ 3 } \)
\(or\quad \frac { 2\pi }{ T } t=\frac { \pi }{ 3 } \)
\(\therefore \quad t=\frac { T }{ 6 } \)

Length of minute hand = radius of circle described by minute hand r = 10 cm = 0.1 m.
From 12.00 noon to 12.30 p.m., the tip of minute hand covers a net displacement equal to the diameter of circle. Hence
Displacement \(\overline{AOB}\) = 2 x r = 2 x 0.1 m = 0.2 m
During the same time total distance covered by tip of minute hand = semicircular path ACB = \(\pi r\) = 3.14 x 0.1 = 0.31 m.

Wavelength of sound wave is \(\lambda = \frac{\upsilon }{v} = \frac{332}{1 \times 10^{3}}= 0.332 m\)
Phase difference between the waves arriving at point of observation is 
\(\phi = \frac{2\pi}{\lambda} \Delta x= \frac{2\pi \times 0.83}{0.332}=5 \pi\)
Since phase difference is an odd multiple of π, the interference is destructive.

D = 2 cm = 0.02 m
\(\rho\) = 10³ kg m^-3
\(\eta\) = 0.001 Nm^-2 s = 10^-3 Nm^-2s
Flow of water will be laminar if
N_R = 1000 where N_R is Reynold number
Let v = maximum average velocity
\(\therefore\) Using the relation
 \(N_R=\frac{\rho vD}{\eta}\)
or   \(v=\frac{N_R\eta}{\rho D}=\frac{1000\times 0.001}{1000\times 0.02}\) = 0.5 ms^-1.

\(D=-N\left( \frac { { x }_{ 2 }-{ x }_{ 1 } }{ { n }_{ 2 }-{ n }_{ 1 } } \right) \)
\(\\ [N]=\frac { { N }_{ 0 } }{ { [L }^{ 2 }T] } =[{ L }^{ -2 }{ T }^{ -1 }]\)
\(\\ [D]=\frac { [{ L }^{ -2 }{ T }^{ -1 }L] }{ [{ L }^{ -3 }] } =[{ L }^{ 2 }{ T }^{ -1 }]\)
\(\\ [{ x }_{ 2 }]=[{ x }_{ 1 }]=[L]\ \)
\(\ [{ n }_{ 2 }]=[{ n }_{ 1 }]=\frac { { N }_{ 0 } }{ [{ L }^{ 3 }] } =[{ L }^{ -3 }]\) 

A given mass of water in vapour state has 1.67×10³ times the volume of the same mass of water in liquid state : For a given mass of water molecules, the density is less if volume is large. So the volume of the vapour is 1000/0.6 = 1/(6 × 10 ^-4 ) times larger. If densities of bulk water and water molecules are same, then the fraction of molecular volume to the total volume in liquid state is 1. As volume in vapour state has increased, the fractional volume is less by the same amount, i.e. 6×10^-4 .
This is also the increase in the amount of volume available for each molecule of water. When volume increases by 103 times the radius increases by V1/3 or 10 times, i.e., 10 × 2 Å = 20 Å. So the average distance is 2 × 20 = 40 Å.

The correct options are
B Pendulum will lose 1.2×10⁻⁴ seconds per each second.
D Pendulum will lose 10.368 seconds/day.
Time period of pendulum at 20^∘C
T20=2π√l/g
where l is the length of pendulum at 20^∘C.
Time period at 40^∘C
\(T_{40}=2 \pi \sqrt{\frac{I(1+a \Delta T)}{g}}\)
therefore Change in time period.
\( T_{40}-T_{20}=2 \pi \sqrt{\frac{1(1+a \Delta T)}{g}}-2 \pi \sqrt{\frac{T}{g}} \)
\( =2 \pi \sqrt{ } \frac{T}{g}(\sqrt{1+a \Delta T}-1)\)
\( \Rightarrow \frac{T_{40}-T_2}{T_{20}}=\sqrt{1+a \Delta T}-1\)
Using binomial expansion, \(\sqrt{1+a \Delta T}=1+\frac{1}{2} a \Delta T\)
\(\Rightarrow \frac{\mathrm{T}_{40}-\mathrm{T}_2 \mathrm{Q}}{\mathrm{T}_{20}}=\frac{1}{2} \mathrm{a} \Delta \mathrm{T}\)
Since the time period increases, pendulum will lose time.
Loss of time per 1 second
\( =\frac{1}{2} \times 0.000012 \times 20=1.2 \times 10^{-4} \mathrm{~s} \)
\( \text { Loss in } 24 \mathrm{hr}=1.2 \times 10^{-4} \times 60 \times 60 \times 24  =10.368 \mathrm{~s} \)

Shear force needed to rupture a bolt
F= Stress×(A)
⇒F=(4.00×10⁸ N/m²) ×π(5.00×10⁻³ m)²
=3.14×10⁴ N
3.14 x 10⁴N, 6.28 x 10⁴N

\(T =2 \pi \sqrt{\frac{R^3}{G M}}=2 \pi \sqrt{\left\{\frac{R^3}{\left(G \frac{4}{3} \pi R^3 \rho\right)}\right\}}=\sqrt{\frac{3 \pi}{G \rho}} \)
\(=\sqrt{\frac{3 \times 3.14}{6.67 \times 10^{-11} \times 8000}} \approx 4202 s \)

Net force is given by Fnet =\(\frac { dp }{ dt } \)
Also, rate of change of momentum = slope of graph.
As from graph, slope AB= slope CD
And slope (BC) = slope (DE) =0
So, force acting on the particle is equal in regions AB and CD and in regions BC and DE (which is zero).

For a projectile launched with velocity \(v _{ 0 }\)at an angle \(\theta _{ 0 }\), the range is given by \(R=\frac { { v }_{ 0 }^{ 2 }sin2\theta _{ 0 } }{ g } \).
Now, for angles, (45° + α) and ( 45° – α), 2θ_o is (90° + 2α) and ( 90° – 2α) , respectively. The values of sin (90° + 2α) and sin (90° – 2α) are the same, equal to that of cos 2α. Therefore, ranges are equal for elevations which exceed or fall short of 45° by equal amounts α.

Here, u = 10.0\(\overset { \wedge }{ j } \) ms^-1 at t = 0.
 (a)\(a=\frac { dv }{ dt } =(8.0\overset { \wedge }{ i } +2.0\overset { \wedge }{ j } ){ ms }^{ -2 }\)
\(\\ so,\quad dv=(8.0\overset { \wedge }{ i } +2.0\overset { \wedge }{ j } )dt\)
Integrating it within the conditions of motion i.e. as time changes from 0 to t, displacement is from 0 to r, we have
\(r=ut+\frac { 1 }{ 2 } \times 8.0{ t }^{ 2 }\overset { \wedge }{ i } +\frac { 1 }{ 2 } \times 2.0{ t }^{ 2 }\overset { \wedge }{ j }\)
\( \\ or\quad x\overset { \wedge }{ i } +y\overset { \wedge }{ j } =10\overset { \wedge }{ j } t+4.0{ t }^{ 2 }\overset { \wedge }{ i } +{ t }^{ 2 }\overset { \wedge }{ j }\)
\( \\ =4.0{ t }^{ 2 }\overset { \wedge }{ i } +)(10t+{ t }^{ 2 })\overset { \wedge }{ j } \)
\(\\ Here, \ we \ have, \ x=4.0{ t }^{ 2 } \ and \ y=10t+{ t }^{ 2 }\)
\(\\ \therefore \quad t={ (x/4) }^{ \frac { 1 }{ 2 } }\)
At x = 16m, t = (16/4)^1/2 = 2s
y = 10 x² + 2²= 24m
(b) Velocity of particle at t= 2 s along x-axis
\(v_{x}=u_{x}+a_{x} t=0+8.0 \times 2=16.0 \mathrm{~m} / \mathrm{s}\)
and along y-axis  \(v_{y}=u_{y}+a_{y} t=10.0+2.0 \times 2=14.0 \mathrm{~m} / \mathrm{s}\)
\(\therefore\) Speed of particle at t = 2s
\(v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{(16.0)^{2}+(14.0)^{2}}=21.26 \mathrm{~ms}^{-1}\)