Have a query? Ask a Tutor!!

Access to best educators and exclusive paper discussions

Offsite Campus Turito Championship

Can’t find what you’re looking for?

Our tutors are from top schools around the world, and they are waiting for you 24/7, anytime, anywhere.

Homework- One to One Solutions
Understand concepts to solve better
Get your doubts solved instantly

Physics-

General

Easy

Question

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

remain same
increase
decrease
may increase or decrease

The correct answer is: decrease

Related Questions to study

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

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

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

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: 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

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

parallel

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

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

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

parallel

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

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

If the time period of a pendulum is 1 sec, then what is the length of the pendulum at point of $l minus T$ and $l minus T to the power of 2 end exponent$ intersection of graph

If the time period of a pendulum is 1 sec, then what is the length of the pendulum at point of $l minus T$ and $l minus T to the power of 2 end exponent$ intersection of graph

physics-General

parallel

The potential energy of a particle of mass 1kg in motion along x-axis is given by $U equals 4 left parenthesis 1 minus c o s invisible function application 2 x right parenthesis$ joule, where ' x' is in meter The period of small oscillations in second is

The potential energy of a particle of mass 1kg in motion along x-axis is given by $U equals 4 left parenthesis 1 minus c o s invisible function application 2 x right parenthesis$ joule, where ' x' is in meter The period of small oscillations in second is

physics-General

For the given figure, calculate zero correction.

For the given figure, calculate zero correction.

physics-General

If ABCD is a square as shown in the figure then which of the following are adjacent sides ?

If ABCD is a square as shown in the figure then which of the following are adjacent sides ?

parallel

card img

With Turito Academy.

card img

With Turito Foundation.

card img

Get an Expert Advice From Turito.

Turito Academy

card img

With Turito Academy.

Test Prep

card img

With Turito Foundation.