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Physics-

General

Easy

Question

Two rods P and Q have equal lengths. Their thermal conductivities are K₁ and K₂and cross sectional areas are A₁ and A₂. When the temperature at ends of each rod are T₁ and T₂ respectively, the rate of flow of heat through P and Q will be equal, if

$\frac{A_{1}}{A_{2}} = \frac{K_{2}}{K_{1}}$
$\frac{A_{1}}{A_{2}} = \frac{K_{2}}{K_{1}} \times \frac{T_{2}}{T_{1}}$
$\frac{A_{1}}{A_{2}} = \sqrt{\frac{K_{1}}{K_{2}}}$
$\frac{A_{1}}{A_{2}} = {(\frac{K_{2}}{K_{1}})}^{2}$

The correct answer is: $fraction numerator A subscript 1 end subscript over denominator A subscript 2 end subscript end fraction equals fraction numerator K subscript 2 end subscript over denominator K subscript 1 end subscript end fraction$

$open parentheses fraction numerator increment Q over denominator increment t end fraction close parentheses subscript P end subscript equals open parentheses fraction numerator increment Q over denominator increment t end fraction close parentheses subscript Q end subscript$
$K subscript 1 end subscript A subscript 1 end subscript fraction numerator open parentheses T subscript 1 end subscript minus T subscript 2 end subscript close parentheses over denominator l end fraction equals K subscript 2 end subscript A subscript 2 end subscript fraction numerator open parentheses T subscript 1 end subscript minus T subscript 2 end subscript close parentheses over denominator l end fraction$
Or $K subscript 1 end subscript A subscript 1 end subscript equals K subscript 2 end subscript A subscript 2 end subscript blank o r blank fraction numerator A subscript 1 end subscript over denominator A subscript 2 end subscript end fraction equals fraction numerator K subscript 2 end subscript over denominator K subscript 1 end subscript end fraction$

Related Questions to study

The graph shows the variation of temperature $left parenthesis T right parenthesis$ of one $k i l o g r a m$ of a material with the heat $left parenthesis H right parenthesis$ supplied to it. At $O comma$ the substance is in the solid state. From the graph, we can conclude that

The graph shows the variation of temperature $left parenthesis T right parenthesis$ of one $k i l o g r a m$ of a material with the heat $left parenthesis H right parenthesis$ supplied to it. At $O comma$ the substance is in the solid state. From the graph, we can conclude that

physics-General

In the adjoining figure, AP and BP are angle bisector of $angle A$ and $angle B$ which meets at P on the parallelogram ABCD. Then $2 angle A P B$ =

Hence option 1 is correct

In the adjoining figure, AP and BP are angle bisector of $angle A$ and $angle B$ which meets at P on the parallelogram ABCD. Then $2 angle A P B$ =

Hence option 1 is correct

A block of mass m is attached with a massless spring of force constant k. The block is placed over a rough inclined surface for which the coefficient of friction is $mu equals 3 divided by 4$ . The minimum value of M required to move the block up the plane is:(Neglect mass of string and pulley and friction in pulley)

A block of mass m is attached with a massless spring of force constant k. The block is placed over a rough inclined surface for which the coefficient of friction is $mu equals 3 divided by 4$ . The minimum value of M required to move the block up the plane is:(Neglect mass of string and pulley and friction in pulley)

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In the figure, a block slides along a track from one level to a higher level, by moving through an intermediate valley. The track is frictionless untill the block reaches the higher level. There a frictional force stops the block in a distance d. The block's initial speed $v subscript 0 end subscript text is end text 6 m divided by s text , end text$ the height difference h is 1.1 m and the coefficient of kinetic friction $mu$ is 0.6. The value of d is

In the figure, a block slides along a track from one level to a higher level, by moving through an intermediate valley. The track is frictionless untill the block reaches the higher level. There a frictional force stops the block in a distance d. The block's initial speed $v subscript 0 end subscript text is end text 6 m divided by s text , end text$ the height difference h is 1.1 m and the coefficient of kinetic friction $mu$ is 0.6. The value of d is

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A ball rolls down an inclined plane figure. The ball is first released from rest from P and then later from Q. Which of the following statement is/are correct?

A ball rolls down an inclined plane figure. The ball is first released from rest from P and then later from Q. Which of the following statement is/are correct?

physics-General

A wedge of mass M fitted with a spring of stiffness 'k' is kept on a smooth horizontal surface. A rod of mass m is kept on the wedge as shown in the figure. System is in equilibrium. Assuming that all surfaces are smooth, the potential energy stored in the spring is:

A wedge of mass M fitted with a spring of stiffness 'k' is kept on a smooth horizontal surface. A rod of mass m is kept on the wedge as shown in the figure. System is in equilibrium. Assuming that all surfaces are smooth, the potential energy stored in the spring is:

physics-General

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A particle 'A' of mass $fraction numerator 10 over denominator 7 end fraction$ kg is moving in the positive x–direction. Its initial position is x = 0 & initial velocity is 1 $m divided by s$ . The velocity at $x equals 10 m$ is:(use the graph given)

A particle 'A' of mass $fraction numerator 10 over denominator 7 end fraction$ kg is moving in the positive x–direction. Its initial position is x = 0 & initial velocity is 1 $m divided by s$ . The velocity at $x equals 10 m$ is:(use the graph given)

physics-General

The perimeter of a right triangle is 132 cm. and the sum of the squares of its sides is 6050 cm². What is the sum of the perpendicular sides –

Hence Choice 1 is correct

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Hence Choice 1 is correct

In figure, triangle ABC is right-angled at B. Given that AB = 9 cm, AC = 15 cm, calculate the length of BC

For such questions, we should know the Pythagoras theorem of a triangle.

In figure, triangle ABC is right-angled at B. Given that AB = 9 cm, AC = 15 cm, calculate the length of BC

For such questions, we should know the Pythagoras theorem of a triangle.

parallel

In DABC, AD is the median through A and E is the produced meets in F figure. Then the value of AF is equal to

Hence AF=(1/3)AC

In DABC, AD is the median through A and E is the produced meets in F figure. Then the value of AF is equal to

Hence AF=(1/3)AC

In the figure shown below, $angle B equals angle C equals 65 to the power of ring operator end exponent$ and $angle D equals 35 to the power of ring operator end exponent$ . Which one of the following is a true statement?

Hence Choice 1 is correct

In the figure shown below, $angle B equals angle C equals 65 to the power of ring operator end exponent$ and $angle D equals 35 to the power of ring operator end exponent$ . Which one of the following is a true statement?

Hence Choice 1 is correct

One of the angles of a triangle is $65 to the power of ring operator end exponent$ . If the difference of other two angles is $25 to the power of ring operator end exponent$ , the largest angle of this triangle has a measure of

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parallel

In figure $angle X equals 62 to the power of ring operator end exponent comma angle X Y Z equals 54 to the power of ring operator end exponent$ If YO and ZO are bisectors of $angle X Y Z$ and $angle X Z Y$ respectively of $capital delta X Y Z$ , find $angle Y O Z$

Hence choice 4 is correct

In figure $angle X equals 62 to the power of ring operator end exponent comma angle X Y Z equals 54 to the power of ring operator end exponent$ If YO and ZO are bisectors of $angle X Y Z$ and $angle X Z Y$ respectively of $capital delta X Y Z$ , find $angle Y O Z$

Hence choice 4 is correct

In figure if $Q T perpendicular P R comma angle T Q R equals 40 to the power of ring operator end exponent$ and $angle S P R equals 30 to the power of ring operator end exponent$ , find x and y.

Hence choice 2 is correct

In figure if $Q T perpendicular P R comma angle T Q R equals 40 to the power of ring operator end exponent$ and $angle S P R equals 30 to the power of ring operator end exponent$ , find x and y.

Hence choice 2 is correct

In Figure. PS is the bisector of $angle Q P R$ and $P T perpendicular Q R$ . Then $angle T P S$ is equal to

Hence Choice 2 is correct

In Figure. PS is the bisector of $angle Q P R$ and $P T perpendicular Q R$ . Then $angle T P S$ is equal to

Hence Choice 2 is correct

parallel

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