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General
Easy
Question
In the experiment to determine Young’s modulus of the material of a wire under tension used in the arrangement as shown The percentage error in the measurement of length is ‘a’, in the measurement of the radius of the wire is ‘b’ and in the measurement of the change in length of the wire is ‘c’ Percentage error in the measurement of Young’s modulus for a given load is
The correct answer is:
Related Questions to study
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The graph shows the change ' ' in the length of a thin uniform wire used by the application of force ‘F’ at different temperatures T1 and T2 The variation suggests that
The graph shows the change ' ' in the length of a thin uniform wire used by the application of force ‘F’ at different temperatures T1 and T2 The variation suggests that
physics-General
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The load versus extension graph for four wires of same material is shown The thinnest wire is represented by the line
The load versus extension graph for four wires of same material is shown The thinnest wire is represented by the line
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Statement–I : As wind flows left to right and a ball is spined as shown, there will be a lift of the ball.
Statement–II : Decrease in velocity of air below the ball, increases the pressure more than that above the ball
Statement–I : As wind flows left to right and a ball is spined as shown, there will be a lift of the ball.
Statement–II : Decrease in velocity of air below the ball, increases the pressure more than that above the ball
physics-General
physics-
Newton's laws of motion can be applied to a block in liquid also. Force due to liquid (e.g., upthrust) are also considered in addition to other forces. A small block of weight W is kept inside. The block is attached with a string connected to the bottom of the vessel. Tension in the string is W/2 If weight of the block is doubled, then tension in the string becomes x times and the time calculated above becomes y times. Then
Newton's laws of motion can be applied to a block in liquid also. Force due to liquid (e.g., upthrust) are also considered in addition to other forces. A small block of weight W is kept inside. The block is attached with a string connected to the bottom of the vessel. Tension in the string is W/2 If weight of the block is doubled, then tension in the string becomes x times and the time calculated above becomes y times. Then
physics-General
physics-
Newton's laws of motion can be applied to a block in liquid also. Force due to liquid (e.g., upthrust) are also considered in addition to other forces. A small block of weight W is kept inside. The block is attached with a string connected to the bottom of the vessel. Tension in the string is W/2 The string is cut. Find the time when it reaches the surface of the liquid
Newton's laws of motion can be applied to a block in liquid also. Force due to liquid (e.g., upthrust) are also considered in addition to other forces. A small block of weight W is kept inside. The block is attached with a string connected to the bottom of the vessel. Tension in the string is W/2 The string is cut. Find the time when it reaches the surface of the liquid
physics-General
physics-
In a U–tube, if different liquids are filled then we can say that pressure at same level of same liquid is same. If small but equal lengths of liquid –1 and liquid –2 are increased in their corresponding sides then h will
In a U–tube, if different liquids are filled then we can say that pressure at same level of same liquid is same. If small but equal lengths of liquid –1 and liquid –2 are increased in their corresponding sides then h will
physics-General
physics-
In a U–tube, if different liquids are filled then we can say that pressure at same level of same liquid is same. In a U–tube 20 cm of a liquid of density 
is on left hand side and 10 cm of another liquid of density 1.5 is on right hand side. In between them there is a third liquid of density 2 . What is the value of h
In a U–tube, if different liquids are filled then we can say that pressure at same level of same liquid is same. In a U–tube 20 cm of a liquid of density is on left hand side and 10 cm of another liquid of density 1.5 is on right hand side. In between them there is a third liquid of density 2 . What is the value of h
physics-General
physics-
When jet of liquid strikes a fixed or moving surface, it exerts thrust on it due to rate of change of momentum. F = 
If surface is free and starts moving due to thrust of liquid, then at any instant, the above equation gets modified based on relative change of momentum with respect to surface. Let any instant the velocity of surface is u, then above equation becomes – 
Based on above concept, in the below given figure, if the cart is frictionless and free to move in horizontal direction, then answer the following :Given cross–section area of jet 
velocity of jet 
= 10 m/s, density of liquid = 1000 kg/ 
, Mass of cart M = 10 kg: The power supplied to the cart, when its velocity becomes 5 m/s, is equal to :
When jet of liquid strikes a fixed or moving surface, it exerts thrust on it due to rate of change of momentum. F =
If surface is free and starts moving due to thrust of liquid, then at any instant, the above equation gets modified based on relative change of momentum with respect to surface. Let any instant the velocity of surface is u, then above equation becomes –
Based on above concept, in the below given figure, if the cart is frictionless and free to move in horizontal direction, then answer the following :Given cross–section area of jet velocity of jet = 10 m/s, density of liquid = 1000 kg/ , Mass of cart M = 10 kg: The power supplied to the cart, when its velocity becomes 5 m/s, is equal to :
physics-General
physics-
When jet of liquid strikes a fixed or moving surface, it exerts thrust on it due to rate of change of momentum. F =
If surface is free and starts moving due to thrust of liquid, then at any instant, the above equation gets modified based on relative change of momentum with respect to surface. Let any instant the velocity of surface is u, then above equation becomes –
Based on above concept, in the below given figure, if the cart is frictionless and free to move in horizontal direction, then answer the following :Given cross–section area of jet velocity of jet = 10 m/s, density of liquid = 1000 kg/ , Mass of cart M = 10 kg: The time at which velocity of cart becomes 2m/s, is equal to:
When jet of liquid strikes a fixed or moving surface, it exerts thrust on it due to rate of change of momentum. F =
If surface is free and starts moving due to thrust of liquid, then at any instant, the above equation gets modified based on relative change of momentum with respect to surface. Let any instant the velocity of surface is u, then above equation becomes –
Based on above concept, in the below given figure, if the cart is frictionless and free to move in horizontal direction, then answer the following :Given cross–section area of jet velocity of jet = 10 m/s, density of liquid = 1000 kg/ , Mass of cart M = 10 kg: The time at which velocity of cart becomes 2m/s, is equal to:
physics-General
physics-
When jet of liquid strikes a fixed or moving surface, it exerts thrust on it due to rate of change of momentum. F =
If surface is free and starts moving due to thrust of liquid, then at any instant, the above equation gets modified based on relative change of momentum with respect to surface. Let any instant the velocity of surface is u, then above equation becomes –
Based on above concept, in the below given figure, if the cart is frictionless and free to move in horizontal direction, then answer the following :Given cross–section area of jet velocity of jet = 10 m/s, density of liquid = 1000 kg/ , Mass of cart M = 10 kg: In the above problem, what is the acceleration of cart at this instant –
When jet of liquid strikes a fixed or moving surface, it exerts thrust on it due to rate of change of momentum. F =
If surface is free and starts moving due to thrust of liquid, then at any instant, the above equation gets modified based on relative change of momentum with respect to surface. Let any instant the velocity of surface is u, then above equation becomes –
Based on above concept, in the below given figure, if the cart is frictionless and free to move in horizontal direction, then answer the following :Given cross–section area of jet velocity of jet = 10 m/s, density of liquid = 1000 kg/ , Mass of cart M = 10 kg: In the above problem, what is the acceleration of cart at this instant –
physics-General
physics-
When jet of liquid strikes a fixed or moving surface, it exerts thrust on it due to rate of change of momentum. F =
If surface is free and starts moving due to thrust of liquid, then at any instant, the above equation gets modified based on relative change of momentum with respect to surface. Let any instant the velocity of surface is u, then above equation becomes –
Based on above concept, in the below given figure, if the cart is frictionless and free to move in horizontal direction, then answer the following :Given cross–section area of jet velocity of jet = 10 m/s, density of liquid = 1000 kg/ , Mass of cart M = 10 kg: Velocity of cart at t = 10 s. is equal to:
When jet of liquid strikes a fixed or moving surface, it exerts thrust on it due to rate of change of momentum. F =
If surface is free and starts moving due to thrust of liquid, then at any instant, the above equation gets modified based on relative change of momentum with respect to surface. Let any instant the velocity of surface is u, then above equation becomes –
Based on above concept, in the below given figure, if the cart is frictionless and free to move in horizontal direction, then answer the following :Given cross–section area of jet velocity of jet = 10 m/s, density of liquid = 1000 kg/ , Mass of cart M = 10 kg: Velocity of cart at t = 10 s. is equal to:
physics-General
physics-
When jet of liquid strikes a fixed or moving surface, it exerts thrust on it due to rate of change of momentum. F =
If surface is free and starts moving due to thrust of liquid, then at any instant, the above equation gets modified based on relative change of momentum with respect to surface. Let any instant the velocity of surface is u, then above equation becomes –
Based on above concept, in the below given figure, if the cart is frictionless and free to move in horizontal direction, then answer the following :Given cross–section area of jet velocity of jet = 10 m/s, density of liquid = 1000 kg/ , Mass of cart M = 10 kg: Initially (t = 0) the force on the cart is equal to :
When jet of liquid strikes a fixed or moving surface, it exerts thrust on it due to rate of change of momentum. F =
If surface is free and starts moving due to thrust of liquid, then at any instant, the above equation gets modified based on relative change of momentum with respect to surface. Let any instant the velocity of surface is u, then above equation becomes –
Based on above concept, in the below given figure, if the cart is frictionless and free to move in horizontal direction, then answer the following :Given cross–section area of jet velocity of jet = 10 m/s, density of liquid = 1000 kg/ , Mass of cart M = 10 kg: Initially (t = 0) the force on the cart is equal to :
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physics-
Statement–I : A block is immersed in a liquid inside a beaker, which is falling freely. Buoyant force acting on block is zero
Statement–II : In case of freely falling liquid there is no pressure difference between any two points
Statement–I : A block is immersed in a liquid inside a beaker, which is falling freely. Buoyant force acting on block is zero
Statement–II : In case of freely falling liquid there is no pressure difference between any two points
physics-General
physics-
Two men ' A' and ' B' are carrying a uniform bar of length ' L' on their shoulders The bar is held horizontally such that A gets one - fourth load If 'A’ is at one end of the bar, the distance of ‘B’ from that end is
Two men ' A' and ' B' are carrying a uniform bar of length ' L' on their shoulders The bar is held horizontally such that A gets one - fourth load If 'A’ is at one end of the bar, the distance of ‘B’ from that end is
physics-General
physics-
Calculate the force ' F 'that is applied horizontally at the axle of the wheel which is necessary to raise the wheel over the obstacle of height 4m Radius of the wheel is 1m and its mass is 10 kg 
Calculate the force ' F 'that is applied horizontally at the axle of the wheel which is necessary to raise the wheel over the obstacle of height 4m Radius of the wheel is 1m and its mass is 10 kg
physics-General
