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

The temperature drop through a two layer furnace wall is $900 to the power of ring operator end exponent C$ . Each layer is of equal area of cross–section. Which of the following action(s) will result in lowering the temperature $theta$ of the interface?

By increasing the thermal conductivity of outer layer
By increasing the thermal conductivity of inner layer
By increasing thickness of outer layer
By increasing thickness of inner layer

The correct answer is: By increasing thickness of inner layer

Rate of heat flow $fraction numerator K subscript i end subscript A open parentheses 1000 minus theta close parentheses over denominator l subscript i end subscript end fraction equals fraction numerator K subscript 0 end subscript A open parentheses theta minus 100 close parentheses over denominator l subscript 0 end subscript end fraction rightwards double arrow fraction numerator theta minus 100 over denominator 900 end fraction equals fraction numerator 1 over denominator 1 plus fraction numerator K subscript 0 end subscript l subscript i end subscript over denominator K subscript i end subscript l subscript 0 end subscript end fraction end fraction$
Now, we can see that $theta$ can be decreased by increasing thermal conductivity of outer layer (Ko ) and thickness of inner layer $text ( end text l subscript i end subscript text ) end text$ .

Related Questions to study

An irregular rod of same uniform material as shown in figure is conducting heat at a steady rate. The temperature gradient at various sections versus area of cross section graph will be

An irregular rod of same uniform material as shown in figure is conducting heat at a steady rate. The temperature gradient at various sections versus area of cross section graph will be

Physics-General

Figure shows the variation of the internal energy U with density r of one mole of an ideal monatomic gas for thermodynamic cycle ABCA. Here process AB is a part of rectangular hyperbola :-

Figure shows the variation of the internal energy U with density r of one mole of an ideal monatomic gas for thermodynamic cycle ABCA. Here process AB is a part of rectangular hyperbola :-

Physics-General

Given T–P curve for three processes. If initial and final pressure are same for all processes then work done in process 1, 2 and 3 is $W subscript 1 end subscript comma W subscript 2 end subscript & W subscript 3 end subscript$ respectively. Correct order is

Given T–P curve for three processes. If initial and final pressure are same for all processes then work done in process 1, 2 and 3 is $W subscript 1 end subscript comma W subscript 2 end subscript & W subscript 3 end subscript$ respectively. Correct order is

Physics-General

parallel

Suppose 0.5 mole of an ideal gas undergoes an isothermal expansion as energy is added to it as heat Q. Graph shows the final volume $V subscript f end subscript$ versus Q. The temperature of the gas is (use $l$ n 9 = 2 and R= 25/3 J/mol^-K)

Suppose 0.5 mole of an ideal gas undergoes an isothermal expansion as energy is added to it as heat Q. Graph shows the final volume $V subscript f end subscript$ versus Q. The temperature of the gas is (use $l$ n 9 = 2 and R= 25/3 J/mol^-K)

Physics-General

Figure demonstrates a polytropic process (i.e. $P V to the power of n end exponent$ = constant) for an ideal gas. The work done by the gas will be in process AB is

Figure demonstrates a polytropic process (i.e. $P V to the power of n end exponent$ = constant) for an ideal gas. The work done by the gas will be in process AB is

Physics-General

The gas left in the vessel will be 2g of hydrogen i.e. number of moles $blank mu equals fraction numerator 2 over denominator 2 end fraction equals 1$

The gas left in the vessel will be 2g of hydrogen i.e. number of moles $blank mu equals fraction numerator 2 over denominator 2 end fraction equals 1$

Physics-General

parallel

A vessel contains 14 g (7 moles) of hydrogen and 96 g (3 moles) of oxygen at STP. Chemical reaction is induced by passing electric spark in the vessel till one of the gases is consumed. The temperature is brought back to it's starting value 273 K. The pressure in the vessel is-

A vessel contains 14 g (7 moles) of hydrogen and 96 g (3 moles) of oxygen at STP. Chemical reaction is induced by passing electric spark in the vessel till one of the gases is consumed. The temperature is brought back to it's starting value 273 K. The pressure in the vessel is-

Physics-General

A fine steel wire of length 4m is fixed rigidly in a heavy brass frame as shown in figure. It is just taut at $20 to the power of ring operator end exponent C$ . The tensile stress developed in steel wire if whole system is heated to $120 to the power of ring operator end exponent C$ is :–
$text (Given end text open alpha subscript text lras end text end subscript equals 1.8 cross times 10 to the power of negative 5 end exponent degree C to the power of negative 1 end exponent comma alpha subscript text sleel end text end subscript equals 1.2 cross times 10 to the power of negative 5 end exponent blank to the power of ring operator end exponent C to the power of negative 1 end exponent comma Y subscript text steel end text end subscript equals 2 cross times 10 to the power of 11 end exponent N m to the power of negative 2 end exponent comma Y subscript text bas end text end subscript equals 1.7 cross times 10 to the power of 7 end exponent N m to the power of negative 2 end exponent close parentheses$

A fine steel wire of length 4m is fixed rigidly in a heavy brass frame as shown in figure. It is just taut at $20 to the power of ring operator end exponent C$ . The tensile stress developed in steel wire if whole system is heated to $120 to the power of ring operator end exponent C$ is :–
$text (Given end text open alpha subscript text lras end text end subscript equals 1.8 cross times 10 to the power of negative 5 end exponent degree C to the power of negative 1 end exponent comma alpha subscript text sleel end text end subscript equals 1.2 cross times 10 to the power of negative 5 end exponent blank to the power of ring operator end exponent C to the power of negative 1 end exponent comma Y subscript text steel end text end subscript equals 2 cross times 10 to the power of 11 end exponent N m to the power of negative 2 end exponent comma Y subscript text bas end text end subscript equals 1.7 cross times 10 to the power of 7 end exponent N m to the power of negative 2 end exponent close parentheses$

Physics-General

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

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

Physics-General

parallel

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

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.