Please wait...
1800-212-7858 (Toll Free)
9:00am - 8:00pm IST all days
For Business Enquiry


Thanks, You will receive a call shortly.
Customer Support

You are very important to us

For any content/service related issues please contact on this toll free number


Mon to Sat - 11 AM to 8 PM

A test tube loaded with lead shots floats to the mark X in water.  The test-tube along with lead shots weighs 25 g.  When the test-tube is floated in brine, 5 g of lead shots were added to make it float upto level X.  Find the relative density of brine.

Asked by Topperlearning User 4th June 2014, 1:23 PM
Answered by Expert

While floating in water, the weight of the floating test tube = 25 gf.  This will be equal to the weight of the water displaced by the floating test-tube, immersed to the mark X.

While floating in brine, the weight of the floating test-tube = 25 + 5 = 30 gf.  Since the test-tube remains immersed to the same mark X, while floating in brine.

Therefore, 30gf is the weight of the same volume of brine.

Since R.D =

Therefore, R.D of brine = 1.2

Answered by Expert 4th June 2014, 3:23 PM
Rate this answer
  • 1
  • 2
  • 3
  • 4
  • 5
  • 6
  • 7
  • 8
  • 9
  • 10

You have rated this answer /10

Your answer has been posted successfully!

Chat with us on WhatsApp