Question

$\large\underline{\sf{ \star \: Question \: 1}}$
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. Taking g=10m s ^{ - 1} and speed of sound =330m s ^{ - 1}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.
$\large\underline{\sf{ \star Question \: 2}}$
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?
$\large\red{\bf{no \: spam \times }}$
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Answer 1

Questions :-

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. Taking g=10m s ^{ - 1} and speed of sound =330m s ^{ - 1}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.

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?

Solutions :-

$\rightarrow$ Question 1:-

(i) Let us assume that the depth of the water surface is "h"

• To reach the water surface, the time taken by the stone = time of                          descent = $\sf\sqrt{\frac{2h}{g} }$ = 20 (given)

$\rightarrow \sf\sqrt{\frac{2h}{g} } = 20$

$\rightarrow \sf\frac{2h}{10} = 400$

$\sf\rightarrow h = 2000 m$    

∴ Depth of water surface is 2000 m

(ii) To hear the echo, the stone has to hit the water surface first and then the sound should travel back to the top

→ For the stone to hit the surface of water, it takes time t₁ = $\sf\sqrt{\frac{2h}{g} }$ and for the sound to travel back to the surface, it takes time t₂ = $\sf\frac{h}{v_{sound}}$ . So total time taken taken to hear the echo is t = t₁ + t₂ = $\sf\sqrt{\frac{2h}{g} } + \sf\frac{h}{v_{sound}}$

$\rightarrow$ t = $\sf\sqrt{\frac{2(2000)}{10} }$ + $\sf\frac{2000}{330}$

$\rightarrow$ t = 20 + 6.06 = 26.06 s

∴ The echo is heard after 26.06 ≈ 26s

$\rightarrow$ Question 2 :-

→ P₁ = 760 mm of Hg

→ V₁ = 850 ml

→ P₂ = ?

→ V₂ = 75 % of initial value = $\sf\frac{75}{100} \times 850$ = 637.5 ml

We know, according to boyle's law, P₁V₁ = P₂V₂

$\rightarrow$ 760 × 850 = P₂ × 637.5

$\rightarrow$ P₂ = 1013.33 mm of Hg

∴ The final pressure of the gas is 1013.33 mm of Hg

_________________________________

Thank you, please mark as brainliest!

$\rightarrow$ Sathvik :)

Answer 2

Answer 1 :

H = 1 × 10 × (20) 2

2

= 5 × 490 = 2099m

Now, t1 = time by falling = 20 sec

= H = 2000

V sonic 330

=> 6.06s

∴ Time after dropping the pebble at which echo is heard =26.06sec

Answer 2 :

P1 = 760mm

V1 = 850ml

V2 = 65100 × 850 = 75 × 8510

P2 = ?

P1 V1 = P2 V2

∴ 760 × 850 = P2 × 75 × 8510

∴ 760 × 10 × 1075 = P2

P2 = 1013.33mm