Minimum speed is at the highest point, since the vertical component of velocity is zero.

 \(\frac{\pi}{2}\) radians.

Inertia of linear motion is measured by mass of the body, which is a scalar quantity.

(i) Kinetic energy does not remain conserved.
(ii) Linear momentum of the system remains conserved.

No. This is because the sounds produced will not have a constant phase difference.

Thermal conductivity, electrical conductivity and compressibility.

Surface tension of the liquid decreases.

When either the lift is at rest or the lift is moving either vertically upward or downward with a constant speed, we can apply three simple kinetnatic motion equations presuming a = ± g (as the case may be).
In present case u = 49 ms^-1 (upward) a = g = 9.8 ms^-2 (downward)
If the ball returns to boy's hands after a time t, then displacement of ball relative to boy is zero
i.e., s = O. Hence, using equation s = ut + \(\frac{1}{2}\) at², we have
0 = 49 ± - \(\frac {1} {2}\) x 9.8 x t^{2
\(\Rightarrow\)} 4.9 t² - 49t = 0 \(\Rightarrow\) t = 0 or 10 s
As t = 0 is physically not possible, hence time t = 10 s.

The coefficient of increase of pressure (or volume) of a gas is It is very large as compared to coefficient of expansion of mercury. Therefore, for a certain increase in volume of the gas will be compared to that of mercury and hence a gas thermometer is more sensitive.

The safety pin is made of steel which is good conductor of heat. So, the safety pin takes heat from the paper under it and transfer it away to the surroundings. The portion of the paper under the safety pin remains comparatively colder than the remaining part.

The critical speed of a fluid depends on ( a ) diameter of tube,  ( b ) density of fluid, ( c ) coefficient of viscosity of the fluid.

\(F=G\frac { { m }_{ 1 }{ m }_{ 2 } }{ { r }^{ 2 } }\)
\( \\ =6.67\times 10^{ -8 })\left( \frac { 1\times 1 }{ 1^{ 2 } } \right) dyne\)
\(\\ =6.67\times 10^{ -8 }\quad dyne=\frac { 6.67\times 10^{ -8 } }{ 980 } \)
\(=7\times 10^{ -11 }g-wt\)

for maximum horizontal range, θ = 45º, Velocity at highest point = Horizontal component of velocity = ucos 45º = u /√2

Wavelength [λ]=[L]
Frequency [ν]=[T⁻¹]

321.14

We know, the vertical distance travelled by a projectile in time t is given by,
Y = u sin \(\theta\)\(\times\)t- \(\frac{1}{2}\)gt²
If h be the height of point p, then for y = It,
we have h = u sin  \(\theta\)\(\times\)t-\(\frac{1}{2}\)gt²
or \(\frac{1}{2}\)gt² - u sin  \(\theta\)\(\times\)t + h = 0
or t²-\(\frac { 2u\quad sin\theta }{ g } .t+\frac { 2h }{ g } \)
This equation is quadratic in t and has two roots t₁ and t₂. Thus there are two values of time for which the height of the projectile is same during flight of projectile.
\(t=\frac { \frac { 2u\quad sin\theta }{ g } \pm \sqrt { \left( \frac { 2u\ sin\theta }{ g } \right) ^{ 2 }-\frac { 8h }{ g } } }{ 2 } \)
\(t=\ \frac { usin\theta }{ g } \pm \sqrt { \frac { { u }^{ 2 }{ sin }^{ 2 }\theta }{ { g }^{ 2 } } -\frac { 2h }{ g } } \)
\(\therefore { t }_{ 1 }=\frac { usin\theta }{ g } \pm \sqrt { \frac { { u }^{ 2 }{ sin }^{ 2 }\theta }{ { g }^{ 2 } } -\frac { 2h }{ g } } \)
and \({ t }_{ 2 }=\frac { usin\theta }{ g } -\sqrt { \frac { { u }^{ 2 }{ sin }^{ 2 }\theta }{ { g }^{ 2 } } -\frac { 2h }{ g } } \)
Now \({ t }_{ 1 }+{ t }_{ 2 }=\frac { 2usin\theta }{ g } \)(Time during the course of flight).
Thus, the sum of times for the same height is equal to the total time of flight.

The situation is shown in Fig
(a) We know that \(\vec { L } =\vec { r } \times m\vec { v } \)
 L= m

Compressibility of the material of a body is defined as the reciprocal of its bulk modulus. It is, thus, defined as the fractional change in volume per unit increase in pressure.
Compressibility, K=\(\frac { 1 }{ B } =-\left( \frac { \triangle V }{ V } \right) \times \frac { 1 }{ P } \)
The solids are least compressible whereas gases are most compressible. It is on account of the fact that in solids neighbouring atoms are tightly coupled but molecules in gases are very poorly coupled to their neighbours.

If two bodies made of same material and have the same mass but different temperatures T₁; and T₂ ' then their equilibrium temperature T when they are brought in thermal contact will be \(\frac { \left( { T }_{ 1 }+{ T }_{ 2 } \right) }{ 2 } \)
[\(\because\) Heat lost = Heat gained]
The equilibrium temperature T will not be necessarily \(\frac { \left( { T }_{ 1 }+{ T }_{ 2 } \right) }{ 2 } \) if two bodies in thermal equilibrium have different heat capacities

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 ana 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 acse which it is assumed \(sin\theta \simeq \theta \)               

Given, T = 273 + 127 = 400 K
Average K.E. per mole of Helium = \(\frac{3}{2} \mathrm{RT}\) 
Average K.E. of I gram of Helium = \(\frac{3}{2} \frac{\mathrm{RT}}{\mathrm{M}}=\frac{3 \times 8.31 \times 400}{2 \times 4}\) 
= 1246.5 J.

\(-6.84\times 10^{ 7 }Jkg^{ -1 }\)

Volume of block, V-lbh
The percentage error in the volume is given by
\(\frac { \triangle V }{ V } \times 100=\left( \frac { \triangle l }{ l } +\frac { \triangle b }{ b } +\frac { \triangle h }{ h } \right) \times 100\)
\(=\left( \frac { 0.02 }{ 12.13 } +\frac { 0.01 }{ 8.16 } +\frac { 0.01 }{ 3.46 } \right) \times 100\)
\(=\frac { 200 }{ 1213 } +\frac { 100 }{ 816 } +\frac { 100 }{ 346 } \)
\(=0.1649+0.1225+0.2890\)
\( =0.58%\)

We first find when and where the two balls collide. Let them collide at an instant f seconds after they start their respective motion. Clearly the two balls are at the same height above the ground at this instant.

The height of the first ball after t seconds = 49 t - 1/2 x 9.8t² = 4.9 t (10 - t)
Also the height of the second ball after t seconds = 98 - downward distance moved by it in t seconds.
= \(98-\frac{1}{2}\times 9.8 t^{2}=4.9(20-t^{2})\)
∴ \(4.9 t (100-t)=4.9(20-t^{2}) \)
or \(10t-t^{2}=20-t^{2}\ or \ t=2s\)
The balls thus collide two seconds after the start of their motion. Their velocities at this instant are
First ball: \(v_{1}=(49-9.8 \times 2) m/s\)
= 29.4 m/ s directed upwards
Second ball: v₂ = \((0+9.8 \times 2)\) m/s
= 19.6 m/s directed downwards
if v is the velocity of the combined mass of the two balls after they stick together folluwing 
their collision, we have, by principle of conservation of momentum.
\(200\times v=100 \times 29.4 -100 \times 19.6\)
∴ v = 4.9 m/s
The 'combined mass' thus moves upward, after collision with a velocity of 4.9 m/s. Its height above the ground at this instant is (considering the position of either of the two balls before collision)
\((98-\frac{1}{2}\times 9.8 \times 2^{2})m= (98-19.6)m = 78.4 m\)
We can now find the time t' taken by the 'combined mass' of the two balls to fall to ground.
We have for this 'combined mass',
u = 4.9 m/s ,s = -78.4 m, a = - g = -9.8 ms^-2
ஃ -78.4= 4.9t^'+1/2 (-9.8)t^'2
or t^'2-t^'-16=0
ஃ \(t^{'}=\frac{1\pm \sqrt{1+64}}{2} = \frac{1\pm 8.06}{2}\)
= 4.532 s (leaving out the negative solution)
The 'combined mass' thus takes 4.53 s to fall to the ground. Since the balls collided
2 s after they started their motion, their total time of flight is (2 + 4.53) s = 6.53 s.

u = 0, a = 2 ms^-2, t = 10s
Using equation, v = u + at, we get
v = 0 + 2 x 10 = 20 ms^-1
(a) Let us first consider horizontal motion. The only force acting on the stone is force of gravity which acts vertically downwards.
Its horizontal component is zero. Moreover, air resistance is to be neglected. So, horizontal motion is uniform motion.
\(\therefore\)v _x = v = 20 ms^-1
Let us now consider vertical motion which is controlled by force of gravity.
u=0, a = g = 10 ms^-2, t = (11 - 10) s = 1s
Using v =u + at, v_y = 0 + 10 x 1 = 10 ms^-1
Resultant velocity,
\(v=\sqrt{V^2_x+V^2_y}\)
\(v=\sqrt{20^2+10^2}ms^{-1}\)
\(=\sqrt{500}ms^{-1}\)
= 22.36 ms^-1

\(tan \beta ={v_y\over v_x}={10\over20}={1\over2}=0.5\)
or \(\beta =\)tan^-1 (0.5) = 26.56°
or \(\beta =\) 26° 34'. This angle is with the horizontal.
(b) The moment the stone is dropped from the car, horizontal force on the stone is zero. The only acceleration of the stone is that due to gravity. This gives a vertically downward acceleration of 10 ms^-2. This is also the net acceleration of the stone.

Bats emit ultrasonic waves of large frequencies. These waves will be reflected by the obstacles in their path. The reflected rays received by the bat will give idea about the obstacle, i.e. distance, direction, size and nature.

Given, T₁ = 27⁰C = 27 + 273 = 300K
T₂ = 0⁰ C = 0 + 273
= 273K
As we know, to freeze one gram of water 0⁰C, 80cal of heat must be transferred from water at 0^oC to the surrounding at 27⁰C.
Q₂ = 80 cal
The coefficient of performance of a refrigerator,
\(\beta =\frac { { Q }_{ 2 } }{ W } =\frac { { T }_{ 2 } }{ { T }_{ 1 }-{ T }_{ 2 } } \Rightarrow \frac { 80 }{ W } =\frac { 273 }{ 300-273 } \)
\(\\ \Rightarrow 7.91\ cal\) 

Let T be the temperature of the hot iron in the furnace. Heat radiated per second per unit area, E = \(\sigma \)T⁴ . When the body is placed in the open at temperature T₀ , the heat radiated/Second/unit area,
E' = \(\sigma \)(T₄-T₀4)
Clearly, E' < E. So, the optical pyrometer gives too low a value for the temperature in the open. 

(i) We employ with the sun’s mass replaced by the martian mass M_m
\({ T^{ 2 }=\frac { { 4\pi }^{ 2 } }{ GM_m } { R }^{ 3 } }\)
\({ M }_{ m }=\frac { { 4\pi }^{ 2 } }{ g } .\frac { { R }^{ 3 } }{ T^{ 2 } }\)
\(=\frac { 4\times (3.14)^{ 2 }\times (9.4)^{ 3 }\times 10^{ 18 } }{ 6.67\times { 10 }^{ -11 }\times (459\times 60)^2 }\)
\(=\frac { 4\times (3.14)^{ 2 }\times (9.4)^{ 3 }\times 10^{ 18 } }{ 6.67\times (459\times 60)^2\times 10^{-5} }\)
\(=6.48\times { 10 }^{ 23 }kg\)
(ii) Once again Kepler’s third law comes to our aid,
\(\frac { { T }_{ m }^{ 2 } }{ { T }^{ 2 } } =\frac { { R }_{ MS }^{ 3 } }{ { R }_{ ES }^{ 3 } } \)
Where R_MS is the mass-sun distance and R_ES is the earth-sun distance
\(\therefore \ \ \ { T }_{ m }=(1.52)^{ 3/2 }\times 365=648\ days\)
We note that the orbits of all planets except Mercury and Mars are very close to being circular. For example, the ratio of the semiminor to semi-major axis for our Earth is, b/a = 0.99986.

Given, speed of man (v_m) = 4 km/h
Speed of river (v_r) = 3 km
Width of the river (d) = 1 km
     
Time taken by the man to cross the river 
t = \(\frac { Width\ of\ the\ river }{ Speed\ of\ the\ man } =\frac { 1\ km }{ 4\ km/h } =\frac { 1 }{ 4 } \)h =\(\frac { 1 }{ 4 } \times 60\) = 15 min
Distance travelled along the river = v_r\(\times \) t
=     \(3\times \frac { 1 }{ 4 } =\frac { 3 }{ 4 } \quad \)km = \(\frac { 3000 }{ 4 } \)
= 750 m