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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 $equals 2 cross times 10 to the power of negative 4 end exponent text end text m to the power of 2 end exponent$ velocity of jet $V subscript 0 end subscript equals 10 text end text m divided by s e c$ ., density of liquid $equals 1000 text end text k g divided by m subscript blank superscript 3 end superscript comma M a s s$ of cart $equals M equals 10 text end text k g$ .The power supplied to the cart, when its velocity becomes 5 m/sec., is equal to :

100 W
25 W
50 W
200 W

The correct answer is: 50 W

$table row cell F equals 2 rho A open parentheses V subscript 0 end subscript minus u close parentheses to the power of 2 end exponent equals 2 cross times 10 to the power of 3 end exponent cross times 2 cross times 10 to the power of negative 4 end exponent cross times 25 equals 10 N end cell row cell P equals F times u equals 10 cross times 5 equals 50 W end cell end table$

Related Questions to study

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 $equals 2 cross times 10 to the power of negative 4 end exponent text end text m to the power of 2 end exponent$ velocity of jet $V subscript 0 end subscript equals 10 text end text m divided by s e c$ ., density of liquid $equals 1000 text end text k g divided by m subscript blank superscript 3 end superscript comma M a s s$ of cart $equals M equals 10 text end text k g$ .The time at which velocity of cart becomes 2 m/sec, is equal to

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 $equals 2 cross times 10 to the power of negative 4 end exponent text end text m to the power of 2 end exponent$ velocity of jet $V subscript 0 end subscript equals 10 text end text m divided by s e c$ ., density of liquid $equals 1000 text end text k g divided by m subscript blank superscript 3 end superscript comma M a s s$ of cart $equals M equals 10 text end text k g$ .The time at which velocity of cart becomes 2 m/sec, is equal to

Physics-General

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 $equals 2 cross times 10 to the power of negative 4 end exponent text end text m to the power of 2 end exponent$ velocity of jet $V subscript 0 end subscript equals 10 text end text m divided by s e c$ ., density of liquid $equals 1000 text end text k g divided by m subscript blank superscript 3 end superscript comma M a s s$ of cart $equals M equals 10 text end text k g$ .In the above problem, what is the acceleration of cart at this instant –

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 $equals 2 cross times 10 to the power of negative 4 end exponent text end text m to the power of 2 end exponent$ velocity of jet $V subscript 0 end subscript equals 10 text end text m divided by s e c$ ., density of liquid $equals 1000 text end text k g divided by m subscript blank superscript 3 end superscript comma M a s s$ of cart $equals M equals 10 text end text k g$ .In the above problem, what is the acceleration of cart at this instant –

Physics-General

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 $equals 2 cross times 10 to the power of negative 4 end exponent text end text m to the power of 2 end exponent$ velocity of jet $V subscript 0 end subscript equals 10 text end text m divided by s e c$ ., density of liquid $equals 1000 text end text k g divided by m subscript blank superscript 3 end superscript comma M a s s$ of cart $equals M equals 10 text end text k g$ .Velocity of cart at t = 10 sec. is equal to :

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 $equals 2 cross times 10 to the power of negative 4 end exponent text end text m to the power of 2 end exponent$ velocity of jet $V subscript 0 end subscript equals 10 text end text m divided by s e c$ ., density of liquid $equals 1000 text end text k g divided by m subscript blank superscript 3 end superscript comma M a s s$ of cart $equals M equals 10 text end text k g$ .Velocity of cart at t = 10 sec. is equal to :

Physics-General

parallel

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 $equals 2 cross times 10 to the power of negative 4 end exponent text end text m to the power of 2 end exponent$ velocity of jet $V subscript 0 end subscript equals 10 text end text m divided by s e c$ ., density of liquid $equals 1000 text end text k g divided by m subscript blank superscript 3 end superscript comma M a s s$ of cart $equals M equals 10 text end text k g$ . Initially $left parenthesis t equals 0 right parenthesis$ the force on the cart is equal to :

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 $equals 2 cross times 10 to the power of negative 4 end exponent text end text m to the power of 2 end exponent$ velocity of jet $V subscript 0 end subscript equals 10 text end text m divided by s e c$ ., density of liquid $equals 1000 text end text k g divided by m subscript blank superscript 3 end superscript comma M a s s$ of cart $equals M equals 10 text end text k g$ . Initially $left parenthesis t equals 0 right parenthesis$ the force on the cart is equal to :

Physics-General

A cylindrical vessel of 90 cm height is kept filled upto the brim. It has four holes 1, 2, 3, 4 which are respectively at heights of 20cm, 30 cm, 40 cm and 50 cm from the horizontal floor PQ. The water falling at the maximum horizontal distance from the vessel comes from

A cylindrical vessel of 90 cm height is kept filled upto the brim. It has four holes 1, 2, 3, 4 which are respectively at heights of 20cm, 30 cm, 40 cm and 50 cm from the horizontal floor PQ. The water falling at the maximum horizontal distance from the vessel comes from

Physics-General

Two forces $F subscript 1 end subscript$ and $F subscript 2 end subscript$ act on a thin uniform elastic rod placed in space. Force $F subscript 1 end subscript$ acts at right end of the rod and $F subscript 2 end subscript$ acts exactly at centre of the rod as shown (both forces act parallel to length of the rod).
i) $F subscript 1 end subscript$ causes extension of rod while $F subscript 2 end subscript$ causes compression of rod.
ii) $F subscript 1 end subscript$ causes extension of rod and $F subscript 2 end subscript$ also causes extension of rod.
iii) $F subscript 1 end subscript$ causes extension of rod while $F subscript 2 end subscript$ does not change length of rod.
The correct order of True / False in above statements is

Two forces $F subscript 1 end subscript$ and $F subscript 2 end subscript$ act on a thin uniform elastic rod placed in space. Force $F subscript 1 end subscript$ acts at right end of the rod and $F subscript 2 end subscript$ acts exactly at centre of the rod as shown (both forces act parallel to length of the rod).
i) $F subscript 1 end subscript$ causes extension of rod while $F subscript 2 end subscript$ causes compression of rod.
ii) $F subscript 1 end subscript$ causes extension of rod and $F subscript 2 end subscript$ also causes extension of rod.
iii) $F subscript 1 end subscript$ causes extension of rod while $F subscript 2 end subscript$ does not change length of rod.
The correct order of True / False in above statements is

Physics-General

parallel

A rod of length 1000 mm and co-efficient of linear expansion $alpha equals 10 to the power of negative 4 end exponent$ per degree is placed symmetrically between fixed walls separated by 1001 mm. The Young's modulus of the rod is $10 to the power of 11 end exponent N divided by m to the power of 2 end exponent$ . If the temperature is increased by $20 to the power of ring operator end exponent C$ , then the stress developed in therod is (in $N divided by m to the power of 2 end exponent$ ):

A rod of length 1000 mm and co-efficient of linear expansion $alpha equals 10 to the power of negative 4 end exponent$ per degree is placed symmetrically between fixed walls separated by 1001 mm. The Young's modulus of the rod is $10 to the power of 11 end exponent N divided by m to the power of 2 end exponent$ . If the temperature is increased by $20 to the power of ring operator end exponent C$ , then the stress developed in therod is (in $N divided by m to the power of 2 end exponent$ ):

Physics-General

A long capillary tube of mass ' $pi$ ' gm, radius 2mm and negligible thickness, is partially immersed in a liquid of surface tension 0.1 N/m. Take angle of contact zero and neglect buoyant force of liquid. The force required to hold the tube vertically, will be - (g = 10 $m divided by s to the power of 2 end exponent$ )

A long capillary tube of mass ' $pi$ ' gm, radius 2mm and negligible thickness, is partially immersed in a liquid of surface tension 0.1 N/m. Take angle of contact zero and neglect buoyant force of liquid. The force required to hold the tube vertically, will be - (g = 10 $m divided by s to the power of 2 end exponent$ )

Physics-General

Assuming the xylem tissues through which water rises from root to the branches in a tree to be of uniform cross-section find the maximum radius of xylum tube in a 10 m high coconut tree so that water can rise to the top. (surface tension of water = 0.1N/m, Angle of contact of water with xylem tube= $60 to the power of ring operator end exponent$ )

Assuming the xylem tissues through which water rises from root to the branches in a tree to be of uniform cross-section find the maximum radius of xylum tube in a 10 m high coconut tree so that water can rise to the top. (surface tension of water = 0.1N/m, Angle of contact of water with xylem tube= $60 to the power of ring operator end exponent$ )

Physics-General

parallel

A block of mass ‘ M ’ area of cross-section ‘A’ & length ‘ $l$ ’ is placed on smooth horizontal floor. A force ‘ F ’ is applied on the block as shown. If ‘ y’ is young modulus of material, then total extension in the block will be:

A block of mass ‘ M ’ area of cross-section ‘A’ & length ‘ $l$ ’ is placed on smooth horizontal floor. A force ‘ F ’ is applied on the block as shown. If ‘ y’ is young modulus of material, then total extension in the block will be:

Physics-General

In determination of young modulus of elasticity of wire, a force is applied and extension is recorded. Initial length of wire is ‘ 1 m’. The curve between extension and stress is depicted then young modulus of wire will be:

In determination of young modulus of elasticity of wire, a force is applied and extension is recorded. Initial length of wire is ‘ 1 m’. The curve between extension and stress is depicted then young modulus of wire will be:

Physics-General

A space 2.5 cm wide between two large plane surfaces is filled with oil. Force required to drag a very thin plate of area 0.5 $m to the power of 2 end exponent$ just midway the surfaces at a speed of 0.5 m/sec is 1N. The coefficient of viscosity in kg–sec/ $m to the power of 2 end exponent$ is :

A space 2.5 cm wide between two large plane surfaces is filled with oil. Force required to drag a very thin plate of area 0.5 $m to the power of 2 end exponent$ just midway the surfaces at a speed of 0.5 m/sec is 1N. The coefficient of viscosity in kg–sec/ $m to the power of 2 end exponent$ is :

Physics-General

parallel

Shape of the meniscus formed by two liquids when capillaries are dipped in them are shown. In I it is hemispherical where as in II it is flat. Pick correct statement regarding contact angles formed by the liquids in both situations

Shape of the meniscus formed by two liquids when capillaries are dipped in them are shown. In I it is hemispherical where as in II it is flat. Pick correct statement regarding contact angles formed by the liquids in both situations

Physics-General

A capillary of the shape as shown is dipped in a liquid. Contact angle between the liquid and the capillary is $0 to the power of 0 end exponent$ and effect of liquid inside the meniscus is to be neglected. T is surface tension of the liquid, r is radius of the meniscus, g is acceleration due to gravity and $rho$ is density of the liquid then height h in equilibrium is

A capillary of the shape as shown is dipped in a liquid. Contact angle between the liquid and the capillary is $0 to the power of 0 end exponent$ and effect of liquid inside the meniscus is to be neglected. T is surface tension of the liquid, r is radius of the meniscus, g is acceleration due to gravity and $rho$ is density of the liquid then height h in equilibrium is

Physics-General

The figure shows a soap film in which a closed elastic thread is lying. The film inside the thread is pricked. Now the sliding wire is moved out so that the surface area increases. The radius of the circle formed by elastic thread will

The figure shows a soap film in which a closed elastic thread is lying. The film inside the thread is pricked. Now the sliding wire is moved out so that the surface area increases. The radius of the circle formed by elastic thread will

Physics-General

parallel

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