Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

A 2m long tube closed at one end is lowered vertically into water until the closed end is flush with the water surface. See figure below. Calculate the water level height in the tube. (Barometric pressure – 1 atm = 10 m of hydrostatic water head, Temperature = 25 o C, Density of water = 1.00 g/ml. Neglect water vapour pressure). (A) 1.01 m (B) 0.29 m (C) 1.71 m (D) 0.92 m



A 2m long tube closed at one end is lowered vertically into water until the closed end is flush with the water surface. See figure below. Calculate the water level height in the tube. (Barometric pressure –
1 atm = 10 m of hydrostatic water head, Temperature = 25oC, Density of water = 1.00 g/ml. Neglect water vapour pressure).


 


           

 


        (A)    1.01 m


        (B)    0.29 m


        (C)    1.71 m


        (D)    0.92 m


Grade:11

2 Answers

Simran Bhatia
348 Points
7 years ago

(B)

 

Sol.   PV = nRT

 

        V =  . T   Þ V = KT

 

        \      log P = log T + log K

 

        Linear dependence with positive slope

 

At the Y-intercept P/t®0 and P®0 inplying P®0, under such conditions all gases show ideal behavior.

 

saboo
20 Points
2 years ago
Questions is based on Boyle’s law 
before the tube is lowered vertically into water, P * V = 1 atm * (2m * base area of tube) (as volume = base area * height).
after tube is lowered into water, the pressure experienced by gas inside the tube is 
                                                             P_{gas} + P_{liquid (height = 2 - h)} = Patm + P_{liquid height 2m}
                  Now, as 1atm is equivalent to 10 m of height of water we can say
                                                             P_{gas} = Patm + P_{liquid height 2m} - P_{liquid (height = 2 - h)}
                                                            P_{gas} = 1 + \frac{2}{10} - \frac{2-h}{10} in atm
                                                i.e. Pgas = 1.2 – (2 – h) / 10 atm
and volume of gas column = base area * h m 
 
Thus, applying boyle’s law, (PV)_{outside water}=(PV)_{inside water}
                                           1 * 2 * base area = {1.2 – (2 -h)/10} * base area * h
solving for h, we get h = 1.708 m

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free