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Physics-

General

Easy

Question

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

$\frac{1}{\sqrt{5}} s$
$\sqrt{5} s$
$\sqrt{3} s$
$\frac{2}{\sqrt{5}} s$

The correct answer is: $fraction numerator 2 over denominator square root of 5 end fraction s$

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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

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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

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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

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

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

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

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 :

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