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Physics-

General

Easy

Question

Water is filled in a cylindrical container to a height of 3 m. The ratio of the cross–sectional area of the orifice and the beaker is 0.1. The square of the speed of the liquid coming out from the orifice is $open parentheses g equals 10 m divided by s to the power of 2 end exponent close parentheses$

$50 m^{2} s^{2}$
$50.5 m^{2} s^{2}$
$51 m^{2} s^{2}$
$52 m^{2} s^{2}$

The correct answer is: $50 text end text m to the power of 2 end exponent text end text s to the power of 2 end exponent$

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A hemispherical portion of radius R is removed from the bottom of a cylinder of radius R. The volume of the remaining cylinder is V and mass M. It is suspended by a string in a liquid of density $rho$ , where it stays vertical. The upper surface of the cylinder is at a depth h below the liquid surface. The force on the bottom of the cylinder by the liquid is :–

A hemispherical portion of radius R is removed from the bottom of a cylinder of radius R. The volume of the remaining cylinder is V and mass M. It is suspended by a string in a liquid of density $rho$ , where it stays vertical. The upper surface of the cylinder is at a depth h below the liquid surface. The force on the bottom of the cylinder by the liquid is :–

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There is a circular tube in a vertical plane. Two liquids which do not mix and of densities $d subscript 1 end subscript$ and $d subscript 2 end subscript$ are filled in the tube. Each liquid subtends $90 to the power of ring operator end exponent$ angle at centre. Radius joining their interface makes an angle $alpha$ with vertical. ratio $d subscript 1 end subscript divided by d subscript 2 end subscript$ is

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A thin liquid film formed between a U-shaped wire and a light slider supports a weight of $1.5 cross times 10 to the power of negative 2 end exponent N$ (see figure). The length of the slider is 30 cm and its weight negligible. The surface tension of the liquid film is :-

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A jar is filled with two non-mixing ligud is 1 and 2 having densities $rho subscript 1 end subscript$ and $rho subscript 2 end subscript$ , respectively. A solid ball, made of a material of density $rho subscript 3 end subscript$ , is dropped in the jar. It comes to equilibrium in the position shown in the figure. Which of the following is true for $rho subscript 1 end subscript comma rho subscript 2 end subscript$ and $rho subscript 3 end subscript$

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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

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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

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Physics-General

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

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

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

Physics-General

When jet of liquid strikes a fixed or moving surface, it exerts thrust on it due to rate of change of momentum. F $equals open parentheses rho A v subscript 0 end subscript close parentheses v subscript 0 end subscript minus open parentheses rho A v subscript 0 end subscript close parentheses v subscript 0 end subscript c o s invisible function application theta equals rho A v subscript 0 end subscript superscript 2 end superscript left square bracket 1 minus c o s invisible function application theta right square bracket$

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

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

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

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

Physics-General

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