# 1013 - Physics Questions Answers

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**Submit By: MANISH SIR**6 year ago

A 20 cm long test tube (cylindrical) is inverted & pushed vertically down into the water . when the closed end is at water surface, how high has the water risen inside the tube? [p atm =10^5 N/m²]

**Asked By: AMIT DAS**7 year ago

**read solutions ( 1 ) | submit your answer**

**Answer Strategies and trick by Manish sir**(it will help you to solve it by yourself)

Let the height of water in the tube is x cm. then according to PV = constant, calculate the pressure by (20-x) cm. column

then P_{0} + pressure by 20 cm water = pressure by (20-x) cm. column + pressure by x cm water column

**Asked By: ASHISH RANA**7 year ago

**Solved By: NIKHIL VARSHNEY**

**read solutions ( 1 ) | submit your answer**

A body is moving towards north with initial velocity of 13m/s. Its is subjected to a retardation of 2 m/s^{2} towards south.The distance travelled in 7th sec is ??

**Asked By: NIKHIL VARSHNEY**7 year ago

**read solutions ( 3 ) | submit your answer**

**Answer Strategies and trick by Manish sir**(it will help you to solve it by yourself)

displacement of any particular sec. = u+at-(a/2)

here direction of motion is opposite to the direction of acceleration so formula will be u-at+(a/2)

displacement in 7th sec = 0

now we know that direction of motion will become south after 6.5 sec. so distance for last half sec = ut+(1/2)at^{2 }= 1/4

same will be the distance for the first half sec of 7th. so total distance for 7th sec = 1/4 + 1/4 = 1/2