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The two ends of a metal rod are matainted at temperatures 100°C and 110°C. The rate of heat flow in the rod is found to be 4.0 J/s. If the ends are maintained at temperatures 200°C and 210°C, the rate of heat flow will be :

(A)

16.8 J/s

(B)

8.0 J/s

(C)

4.0 J/s

(D)

44.0 J/s

NEET - Medical UG-2015 Medium

Answer & Solution:

Rate of heat flow  \(\large =\frac{d\theta}{dt}=\frac{kA}{L}\left ( T_{1}-T_{2} \right )\)

So rate of heat flow \(\large =\frac{d\theta}{dt} \propto \left ( T_{1}-T_{2} \right )\) 

In this case temperature difference in both \(\large =10^{\circ}C\)

So Heat flow will be same = 4.0 J/s

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Other Questions of the same topic

Q1:

A mass m is attached to a thin wire and whirled in a vertical circle. The wire is most likely to break when:

(A)

 the mass is at the highest point

(B)

the wire is horizontal

(C)

the mass is at the lowest point

(D)

inclined at an angle of 60° from vertical

NEET - Medical UG-2019 Medium

Q2:

A soap bubble, having radius of 1 mm, is blown from a detergent solution having a surface tension of 2.5 × \(\large 10^{-2}\) N/m. The pressure inside the bubble equals at a point \(\large Z_0\) below the free surface of water in a container. Taking g = 10 \(\large m/s^2\), density of water = \(\large 10^3 kg/m^3\), value of \(\large Z_0\) is :

(A)

100 cm

(B)

10 cm

(C)

1 cm

(D)

0.5 cm

NEET - Medical UG-2019 Medium

Q3:

Copper of fixed volume 'V' is drawn into wire of length'\(\large l \)'. When this wire is subjected to a constant force 'F', the extension produced in the wire is \(\large \Delta l \)

;. Which of the following graph is a straight line?

(A)

\(\large \Delta l \: \: versus\: \: 1/l\)

(B)

\(\large \Delta l \: \: versus\: \: l^{2}\)

(C)

\(\large \Delta l \: \: versus\: \:1/ l^{2}\)

(D)

\(\large \Delta l \: \: versus\: \:l\)

NEET - Medical UG-2014 Easy

Q4:

A certain number of spherical drops of a liquid of radius 'r' coalesce to form a single drop of radius 'R' and volume 'V'. If 'T' is the surface tension of the liquid then:

(A)

\(\large Energy= 4VT \left ( \frac{1}{r}-\frac{1}{R} \right )is\: \: released.\)

(B)

\(\large Energy= 3VT \left ( \frac{1}{r}+\frac{1}{R} \right )is\: \: released.\)

(C)

\(\large Energy= 3VT \left ( \frac{1}{r}-\frac{1}{R} \right )is\: \: released.\)

(D)

Energy is neither released nor absorbed.

NEET - Medical UG-2014 Medium

Chemistry Biology

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