Physics-
General
Easy

Question

The load versus extension graph for four wires of same material is shown The thinnest wire is represented by the line

  1. OA    
  2. OB    
  3. OC    
  4. OD    

The correct answer is: OA

Related Questions to study

General
physics-

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
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
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
parallel
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
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 rho is on left hand side and 10 cm of another liquid of density 1.5 rho is on right hand side. In between them there is a third liquid of density 2 rho. 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 rho is on left hand side and 10 cm of another liquid of density 1.5 rho is on right hand side. In between them there is a third liquid of density 2 rho. What is the value of h

physics-General
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 = open parentheses rho A v subscript 0 end subscript close parentheses v subscript 0 end subscript minus open parentheses rho A v subscript 0 end subscript close parentheses v subscript 0 end subscript c o s invisible function application theta equals rho A v subscript 0 end subscript superscript 2 end superscript left square bracket 1 minus c o s invisible function application theta right square bracket

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 – F equals rho A open parentheses v subscript 0 end subscript minus u close parentheses to the power of 2 end exponent left square bracket 1 minus c o s invisible function application theta right square bracket

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 equals 2 cross times 10 to the power of negative 4 end exponent m to the power of 2 end exponent velocity of jet V subscript 0 end subscript = 10 m/s, density of liquid = 1000 kg/ m to the power of 3 end exponent , 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 = open parentheses rho A v subscript 0 end subscript close parentheses v subscript 0 end subscript minus open parentheses rho A v subscript 0 end subscript close parentheses v subscript 0 end subscript c o s invisible function application theta equals rho A v subscript 0 end subscript superscript 2 end superscript left square bracket 1 minus c o s invisible function application theta right square bracket

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 – F equals rho A open parentheses v subscript 0 end subscript minus u close parentheses to the power of 2 end exponent left square bracket 1 minus c o s invisible function application theta right square bracket

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 equals 2 cross times 10 to the power of negative 4 end exponent m to the power of 2 end exponent velocity of jet V subscript 0 end subscript = 10 m/s, density of liquid = 1000 kg/ m to the power of 3 end exponent , Mass of cart M = 10 kg: The power supplied to the cart, when its velocity becomes 5 m/s, is equal to :

physics-General
parallel
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 = open parentheses rho A v subscript 0 end subscript close parentheses v subscript 0 end subscript minus open parentheses rho A v subscript 0 end subscript close parentheses v subscript 0 end subscript c o s invisible function application theta equals rho A v subscript 0 end subscript superscript 2 end superscript left square bracket 1 minus c o s invisible function application theta right square bracket

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 – F equals rho A open parentheses v subscript 0 end subscript minus u close parentheses to the power of 2 end exponent left square bracket 1 minus c o s invisible function application theta right square bracket

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 equals 2 cross times 10 to the power of negative 4 end exponent m to the power of 2 end exponent velocity of jet V subscript 0 end subscript = 10 m/s, density of liquid = 1000 kg/ m to the power of 3 end exponent , 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 = open parentheses rho A v subscript 0 end subscript close parentheses v subscript 0 end subscript minus open parentheses rho A v subscript 0 end subscript close parentheses v subscript 0 end subscript c o s invisible function application theta equals rho A v subscript 0 end subscript superscript 2 end superscript left square bracket 1 minus c o s invisible function application theta right square bracket

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 – F equals rho A open parentheses v subscript 0 end subscript minus u close parentheses to the power of 2 end exponent left square bracket 1 minus c o s invisible function application theta right square bracket

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 equals 2 cross times 10 to the power of negative 4 end exponent m to the power of 2 end exponent velocity of jet V subscript 0 end subscript = 10 m/s, density of liquid = 1000 kg/ m to the power of 3 end exponent , Mass of cart M = 10 kg: The time at which velocity of cart becomes 2m/s, is equal to:

physics-General
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 = open parentheses rho A v subscript 0 end subscript close parentheses v subscript 0 end subscript minus open parentheses rho A v subscript 0 end subscript close parentheses v subscript 0 end subscript c o s invisible function application theta equals rho A v subscript 0 end subscript superscript 2 end superscript left square bracket 1 minus c o s invisible function application theta right square bracket

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 – F equals rho A open parentheses v subscript 0 end subscript minus u close parentheses to the power of 2 end exponent left square bracket 1 minus c o s invisible function application theta right square bracket

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 equals 2 cross times 10 to the power of negative 4 end exponent m to the power of 2 end exponent velocity of jet V subscript 0 end subscript = 10 m/s, density of liquid = 1000 kg/ m to the power of 3 end exponent , 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 = open parentheses rho A v subscript 0 end subscript close parentheses v subscript 0 end subscript minus open parentheses rho A v subscript 0 end subscript close parentheses v subscript 0 end subscript c o s invisible function application theta equals rho A v subscript 0 end subscript superscript 2 end superscript left square bracket 1 minus c o s invisible function application theta right square bracket

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 – F equals rho A open parentheses v subscript 0 end subscript minus u close parentheses to the power of 2 end exponent left square bracket 1 minus c o s invisible function application theta right square bracket

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 equals 2 cross times 10 to the power of negative 4 end exponent m to the power of 2 end exponent velocity of jet V subscript 0 end subscript = 10 m/s, density of liquid = 1000 kg/ m to the power of 3 end exponent , Mass of cart M = 10 kg: In the above problem, what is the acceleration of cart at this instant –

physics-General
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 = open parentheses rho A v subscript 0 end subscript close parentheses v subscript 0 end subscript minus open parentheses rho A v subscript 0 end subscript close parentheses v subscript 0 end subscript c o s invisible function application theta equals rho A v subscript 0 end subscript superscript 2 end superscript left square bracket 1 minus c o s invisible function application theta right square bracket

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 – F equals rho A open parentheses v subscript 0 end subscript minus u close parentheses to the power of 2 end exponent left square bracket 1 minus c o s invisible function application theta right square bracket

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 equals 2 cross times 10 to the power of negative 4 end exponent m to the power of 2 end exponent velocity of jet V subscript 0 end subscript = 10 m/s, density of liquid = 1000 kg/ m to the power of 3 end exponent , 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 = open parentheses rho A v subscript 0 end subscript close parentheses v subscript 0 end subscript minus open parentheses rho A v subscript 0 end subscript close parentheses v subscript 0 end subscript c o s invisible function application theta equals rho A v subscript 0 end subscript superscript 2 end superscript left square bracket 1 minus c o s invisible function application theta right square bracket

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 – F equals rho A open parentheses v subscript 0 end subscript minus u close parentheses to the power of 2 end exponent left square bracket 1 minus c o s invisible function application theta right square bracket

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 equals 2 cross times 10 to the power of negative 4 end exponent m to the power of 2 end exponent velocity of jet V subscript 0 end subscript = 10 m/s, density of liquid = 1000 kg/ m to the power of 3 end exponent , Mass of cart M = 10 kg: Velocity of cart at t = 10 s. is equal to:

physics-General
parallel
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 = open parentheses rho A v subscript 0 end subscript close parentheses v subscript 0 end subscript minus open parentheses rho A v subscript 0 end subscript close parentheses v subscript 0 end subscript c o s invisible function application theta equals rho A v subscript 0 end subscript superscript 2 end superscript left square bracket 1 minus c o s invisible function application theta right square bracket

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 – F equals rho A open parentheses v subscript 0 end subscript minus u close parentheses to the power of 2 end exponent left square bracket 1 minus c o s invisible function application theta right square bracket

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 equals 2 cross times 10 to the power of negative 4 end exponent m to the power of 2 end exponent velocity of jet V subscript 0 end subscript = 10 m/s, density of liquid = 1000 kg/ m to the power of 3 end exponent , 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 = open parentheses rho A v subscript 0 end subscript close parentheses v subscript 0 end subscript minus open parentheses rho A v subscript 0 end subscript close parentheses v subscript 0 end subscript c o s invisible function application theta equals rho A v subscript 0 end subscript superscript 2 end superscript left square bracket 1 minus c o s invisible function application theta right square bracket

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 – F equals rho A open parentheses v subscript 0 end subscript minus u close parentheses to the power of 2 end exponent left square bracket 1 minus c o s invisible function application theta right square bracket

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 equals 2 cross times 10 to the power of negative 4 end exponent m to the power of 2 end exponent velocity of jet V subscript 0 end subscript = 10 m/s, density of liquid = 1000 kg/ m to the power of 3 end exponent , Mass of cart M = 10 kg: Initially (t = 0) the force on the cart is equal to :

physics-General
General
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
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
parallel
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 open parentheses g equals 10 m s to the power of negative 2 end exponent close parentheses

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 open parentheses g equals 10 m s to the power of negative 2 end exponent close parentheses

physics-General
General
physics-

A wheel of radius ' r 'and mass 'm' stands in front of a step of height ' h' The least horizontal force which should be applied to the axle of the wheel to allow it to raise onto the step is

A wheel of radius ' r 'and mass 'm' stands in front of a step of height ' h' The least horizontal force which should be applied to the axle of the wheel to allow it to raise onto the step is

physics-General
General
physics-

A cubical block of side ‘L’ rests on a rough horizontal surface with coefficient of friction A horizontal force F is applied on them' ' block as shown If the coefficient of friction is sufficiently high so that the block does not slide before toppling, the minimum force required to topple the block is

A cubical block of side ‘L’ rests on a rough horizontal surface with coefficient of friction A horizontal force F is applied on them' ' block as shown If the coefficient of friction is sufficiently high so that the block does not slide before toppling, the minimum force required to topple the block is

physics-General
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