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Physics-

General

Easy

Question

The adjacent graph shows the extension $left parenthesis capital delta l$ ) of a wire of length 1m suspended from the top of a roof at one end and with a load W connected to the other end. If the cross–sectional area of the wire is $10 to the power of negative 6 end exponent m to the power of 2 end exponent$ , calculate the Young's modulus of the material of the wire :–

$2 \times 10^{11} N / m^{2}$
$2 \times 10^{- 11} N / m^{2}$
$3 \times 10^{12} N / m^{2}$
$2 \times 10^{13} N / m^{2}$

The correct answer is: $2 cross times 10 to the power of 11 end exponent text end text N divided by m to the power of 2 end exponent$

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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

parallel

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

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

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: Velocity of cart at t = 10 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: 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 $equals open parentheses rho A v subscript 0 end subscript close parentheses v subscript 0 end subscript minus open parentheses rho A v subscript 0 end subscript close parentheses v subscript 0 end subscript c o s invisible function application theta equals rho A v subscript 0 end subscript superscript 2 end superscript left square bracket 1 minus c o s invisible function application theta right square bracket$

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: Initially (t = 0) the force on the cart is equal to :

Physics-General

parallel

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