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Physics-

General

Easy

Question

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

$\frac{1 + t a n α}{1 - t a n α}$
$\frac{1 + s i n α}{1 - c o s α}$
$\frac{1 + s i n α}{1 - s i n α}$
$\frac{1 + \cos α}{1 - \cos α}$

The correct answer is: $fraction numerator 1 plus t a n invisible function application alpha over denominator 1 minus t a n invisible function application alpha end fraction$

Related Questions to study

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 :-

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 :-

Physics-General

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 $rho subscript 3 end subscript$

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 $rho subscript 3 end subscript$

Physics-General

A spherical solid ball of volume V is made of a material of density $rho subscript 1 end subscript$ . It is falling through a liquid of density $stack rho subscript 2 end subscript with bar on top open parentheses rho subscript 2 end subscript less than rho subscript 1 end subscript close parentheses$ . Assume that the liquid applies a viscous force on the ball that is propoertional to the square of its speed v, i.e., $F subscript text viscous end text end subscript equals negative k v to the power of 2 end exponent left parenthesis k greater than 0 right parenthesis$ . Then terminal speed of the ball is

A spherical solid ball of volume V is made of a material of density $rho subscript 1 end subscript$ . It is falling through a liquid of density $stack rho subscript 2 end subscript with bar on top open parentheses rho subscript 2 end subscript less than rho subscript 1 end subscript close parentheses$ . Assume that the liquid applies a viscous force on the ball that is propoertional to the square of its speed v, i.e., $F subscript text viscous end text end subscript equals negative k v to the power of 2 end exponent left parenthesis k greater than 0 right parenthesis$ . Then terminal speed of the ball is

Physics-General

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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

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

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

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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

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 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

Statement–I : For Reynold number $R subscript e end subscript greater than 2000$ , the flow of fluid is turbulent
Statement–II : Inertial forces are dominant compared to the viscous forces at such high Reynold numbers

Statement–I : For Reynold number $R subscript e end subscript greater than 2000$ , the flow of fluid is turbulent
Statement–II : Inertial forces are dominant compared to the viscous forces at such high Reynold numbers

Physics-General

Statement–I : A parachute descends slowly whereas a stone dropped from same height falls rapidly
Statement–II : The viscous force of air on parachute is larger than that of on a falling stone

Statement–I : A parachute descends slowly whereas a stone dropped from same height falls rapidly
Statement–II : The viscous force of air on parachute is larger than that of on a falling stone

Physics-General

In ABC ,AD is median, point F lies on AD such that $fraction numerator A F over denominator F D end fraction equals fraction numerator 1 over denominator 5 end fraction$ _,and $A B & C F$ intersect at E and BF and AC intersect at G.
If $stack a with minus on top comma stack b with minus on top comma stack c with minus on top$ are position vectors of points A, B, C respectively such that $stack a with minus on top comma stack b with minus on top comma stack c with minus on top$ are non-coplanar and if mid-point of AG $lambda stack a with minus on top plus mu stack b with minus on top plus v stack c with minus on top$ , then the value of $lambda plus mu plus v$ is

In ABC ,AD is median, point F lies on AD such that $fraction numerator A F over denominator F D end fraction equals fraction numerator 1 over denominator 5 end fraction$ _,and $A B & C F$ intersect at E and BF and AC intersect at G.
If $stack a with minus on top comma stack b with minus on top comma stack c with minus on top$ are position vectors of points A, B, C respectively such that $stack a with minus on top comma stack b with minus on top comma stack c with minus on top$ are non-coplanar and if mid-point of AG $lambda stack a with minus on top plus mu stack b with minus on top plus v stack c with minus on top$ , then the value of $lambda plus mu plus v$ is

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