Question

★Question 1.)
A pebble is dropped freely in a well from its top. It takes 20 sec for the pebble to reach the water surface in the well.

$\sf \: Taking g=10m s ^{ - 1}  and \: speed \: of \: sound =330m s ^{ - 1}$

find : (i) the depth of water surface, (ii) the time when echo is heard after the pebble is dropped.

★Question 2.)
A certain mass of gas occupied 850ml at a pressure of 760 mm of Hg. On increasing the pressure it was found that the volume of  the gas was 75% of its initial value. Assuming constant temperature, find the final pressure of the gas ?

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Answer 1

1. Given

• A pebble is dropped freely in a well from the top.
• Time taken for the pebble to reach the water surface in the wall = 20 seconds.
• g = 10 m/s²
• Speed of sound = 330 m/s

______________________________

To Find

• The depth of water surface
• The time when the echo is heard after the pebble is dropped.

______________________________

Solution

(i) Initial Velocity (u) = 0 m/s

Time Taken (t) = 20 seconds

Acceleration (a) = 10 m/s²

To find the depth of the water surface, we'll apply the second equation of motion.

$\boxed{ \sf Second \ Equation \ of \ Motion : s = ut + \dfrac{1}{2}at^{2}}$

Substitute the values of the variables and find the value of 's'.

$\implies s = 0(20) + \dfrac{1}{2} \times 10 \times (20)^{2}$

$\implies s = 0 + \dfrac{1}{2} \times 10 \times 400$

$\implies s = 5 \times 400$

$\implies \bf s = 2000 \ m$

∴ The depth of the water surface is 2000 m.

______________________________

(ii) We'll apply the Speed Distance Time Formula.

$\boxed{\sf Time (t) = \dfrac{Distance}{Speed}}$

Modifying this formula,

$\boxed{\sf Time (t) = \dfrac{Depth \ of \ Water \ Surface}{Speed \ of \ Sound}}$

Substitute the values of the variables and find the value of 't'.

$\implies t = \dfrac{2000}{330}$

$\implies t= \dfrac{200}{33}$

$\implies \bf t = 6.06 \ sec$

∴ 6.06 seconds are taken for the pebble to create echo after it reaches the water surface.

Since the pebble takes 20 seconds to reach the water surface in the well, we'll add 6.06 seconds to it.

⇒ Total Time Taken = 20 + 6.06

⇒ Total Time Taken = 26.06 seconds

∴ It would take 26.06 seconds when echo is heard after the pebble is dropped.

______________________________

2.

P₁ = 760 mm

V₁ = 850 ml

P₂ = ?

V₂ = 75% of 850

Let's find the value of V₂.

⇒ 75% × 850

$\implies \dfrac{75}{100} \times 850$

$\implies \dfrac{75}{10} \times 85$

$\implies \dfrac{6375}{10}$

⇒ 637.5 ml

∴ V₂ = 637.5 ml

To find the final pressure of the gas, we'll apply the Boyle's Law.

$\boxed{\sf P_1V_1 = P_2V_2}$

Here,

• P₁ = Initial Pressure
• P₂ = Final Pressure
• V₁ = Initial Volume
• V₂ = Final Volume

⇒ 760 × 850 = 637.5 × x

⇒ 646000 = 637.5x

⇒ x = 646000 ÷ 637.5

⇒ x = 1013.33 mm

∴ The final pressure of the gas would be 1013.33 mm.

______________________________

Answer 2

Question: A pebble is dropped freely in a well from its top. It takes 20 sec for the pebble to reach the water surface in the well. (taking g = 10 m/s², speed of sound = 330 m/s)

find : (i) the depth of water surface, (ii) the time when echo is heard after the pebble is dropped.

Answer:

(i) 2000 m (ii) 26.1 sec

Explanation:

Given: A pebble is dropped freely in a well from its top. It takes 20 sec for the pebble to reach the water surface in the well.

g = 10 m/s², speed of sound = 330 m/s

To find: (i) the depth of water surface,

(ii) the time when echo is heard after the pebble is dropped.

$\rule{40mm}{1pt}$

(i) Let's say that stone is dropped in a well of depth "h" metre.

The required formula is h = ut + 1/2gt²

Here h represents height, u represents initial velocity, t represents time and g represents acceleration.

Value of u is 0 m/s because a freely falling object falls from rest, g is 10 m/s² and t is 20 sec.

$\implies{\sf{h=ut+1/2gt^{2}}}$

$\implies{\sf{h=(0)t+1/2(10)(20)^{2}}}$

$\implies{\sf{h=0+5(400)}}$

$\implies{\sf{h=2000}}$

Therefore, the depth of water surface from the top of the well is 2000 m.

$\rule{40mm}{1pt}$

(ii) Let's say that the time when echo is heard after the pebble is dropped is "T" sec.

Required formula: Time taken to hear the echo after pebble reaches the water surface = (Distance or Depth)/Speed

Substitute the values,

Where the depth water surface from the top of the well is 2000 m (solved above) and speed of sound is 330 m/s (given).

$\implies{\sf{T=2000/330}}$

$\implies{\sf{T=6.1\:sec\:(approx.)}}$

As per given problem time taken by the pebble to reach the water surface is 20 sec and time taken to hear the echo after the pebble reached the surface of water is 6.1 sec.

Therefore,

Total time time echo is heard after the pebble is dropped = 20 sec + 6.1 sec = 26.1 sec

$\rule{70mm}{2pt}$

Question: A certain mass of gas occupied 850ml at a pressure of 760 mm of Hg. On increasing the pressure it was found that the volume of  the gas was 75% of its initial value. Assuming constant temperature, find the final pressure of the gas ?

Answer:

1013.33 mm of Hg

Explanation:

Given that a certain mass of gas occupied 850ml at a pressure of 760 mm of Hg. On increasing the pressure it was found that the volume of  the gas was 75% of its initial value. We need to find out the final pressure of the gas; assuming that temperature is constant.

Now,

There are some laws that describes the behaviour of gases known as gas laws.

1) Boyle's Law: At constant temperature (T), the volume (V) is a given mass of gas is inversely proportional to it's pressure (P).

→ V is inversely proportional to P at constant temperature.

→ P1V1 = P2V2

2) Charles Law: At constant pressure, volume of a given mass of gas is directly proportional to the temperature of gas.

→ V is directly proportional to T, at constant pressure.

→ V1/T1 = V2/T2

3) Gay Lussac's Law: At constant volume, pressure of a given mass of gas is directly proportional to temperature.

→ P1/T1 = P2/T2

4) Avogadro Law: This law states that equal volume of all gases under the same condition of temperature and pressure contain equal number of molecules.

$\rule{40mm}{1pt}$

As per given problem, the law that we need here is Boyle's Law.

$\implies{\sf{P1V1=P2V2}}$

Here P1 is 760 mm, V1 is 850 ml, V2 is 75% of initial value.

V2 = 75/100 × 850 ml

Substitute the values,

$\implies{\sf{760(850)=P2(3*850/4)}}$

$\implies{\sf{760=P2(3/4)}}$

$\implies{\sf{[760(4)]/3=P2}}$

$\implies{\sf{1013.33=P2}}$

Therefore, the final pressure of the gas is 1013.33 mm of Hg.