AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

Two syringes of different cross section (without needle) filled with water are connected with a tightly fitted rubber tube filled with water. Diameters of the smaller piston and larger piston are 1 cm and 3 cm respectively. If a force of 10 N is applied to the smaller piston then the force exerted on the larger piston is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

30 N

60 N

90 N

100 N

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Since pressure is transmitted undiminished through out the water

$\frac{F_{1}}{A_{1}} = \frac{F_{2}}{A_{2}}$

where $F_{1} and F_{2}$ are the forces on the surface and on the larger pistons respectively and $A_{1} and A_{2}$ are the respective areas.

$F_{2} = \frac{A_{2}}{A_{1}} F_{1} = \frac{π {(\frac{D_{2}}{2})}^{2}}{π {(\frac{D_{1}}{2})}^{2}} F_{1} {= (\frac{D_{2}}{D_{1}})}^{2} F_{1}$

= $\frac{{(3 \times 10^{- 2} m)}^{2}}{{(1 \times 10^{- 2} m)}^{2}} \times 10 N = 90 N$

Watch 3-min video & get full concept clarity

courses

No courses found

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!