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

Figures shows the expansion of a 2m long metal rod with temperature. The volume expansion coefficient of the metal is :–

$3 \times 10^{- 4} K^{- 1}$
$1.5 \times 10^{- 4} K^{- 1}$
$3 \times 10^{- 5} K^{- 1}$
$1.5 \times 10^{- 5} K^{- 1}$

The correct answer is: $1.5 cross times 10 to the power of negative 4 end exponent K to the power of negative 1 end exponent$

$alpha equals fraction numerator capital delta l over denominator l capital delta T end fraction equals fraction numerator 0.5 cross times 10 to the power of negative 3 end exponent over denominator 2 open parentheses 5 close parentheses end fraction equals 5 cross times 10 to the power of negative 5 end exponent K to the power of negative 1 end exponent rightwards double arrow gamma equals 3 alpha equals 1.5 cross times 10 to the power of negative 4 end exponent K to the power of negative 1 end exponent$

Related Questions to study

The figure shows two thin rods, one made of aluminum $open square brackets alpha equals 23 cross times 10 to the power of negative 6 end exponent open parentheses C to the power of 0 end exponent close parentheses to the power of negative 1 end exponent close square brackets$ and the other of steel $open square brackets alpha equals 12 cross times 10 to the power of negative 6 end exponent open parentheses C to the power of 0 end exponent close parentheses to the power of negative 1 end exponent close square brackets$ . Each rod has the same length and the same initial temperature. They are attached at one end to two separate immovable walls. Temperature of both the rods is increased by the same amount, until the gap between the rods vanishes. Where do the rods meet when the gap vanishes?

The figure shows two thin rods, one made of aluminum $open square brackets alpha equals 23 cross times 10 to the power of negative 6 end exponent open parentheses C to the power of 0 end exponent close parentheses to the power of negative 1 end exponent close square brackets$ and the other of steel $open square brackets alpha equals 12 cross times 10 to the power of negative 6 end exponent open parentheses C to the power of 0 end exponent close parentheses to the power of negative 1 end exponent close square brackets$ . Each rod has the same length and the same initial temperature. They are attached at one end to two separate immovable walls. Temperature of both the rods is increased by the same amount, until the gap between the rods vanishes. Where do the rods meet when the gap vanishes?

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 $equals M equals 10 text end text k g$ .The power supplied to the cart, when its velocity becomes 5 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 power supplied to the cart, when its velocity becomes 5 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 $equals M equals 10 text end text k g$ .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

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 $equals M equals 10 text end text k g$ .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

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

parallel

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

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

parallel

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

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

parallel

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

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

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.