**“ The pursuit of truth and beauty is a sphere of activity in which we are permitted to remain children all our lives”**

**– Albert Einstein**

As promised in the last post dated 6^{th} February 2009, I give below a few multiple choice practice questions on

**(1) **A soft iron rod has its length increased by 0.3% on increasing its *∆T*. The percentage increase in the area of a circular hole in a sheet of the same material (soft iron) on increasing its *∆T *will be

(a) 0.3%

(b) – 0.3%

(c) 0%

(d) 0.6%

(e) –0.6%

The radius of the hole will *increase* by 0.3% and hence the area of the hole will *increase* by 0.6%.

[The area *A = *π*R*^{2} where *R* is the radius. Therefore, *∆A/A =*2 *∆R/R*. The fractional change in area is equal to *twice* the fractional change in radius. Therefore, the percentage change in area is equal to *twice* the percentage change in radius].

**(2)** The *t*º C and 2*t*º F. The

(a) 32º C

(b) 100º C

(c) 120º C

(d) 142º C

(e) 160º C

We have* **t*_{C}**/100 = ( t_{F }– 32)_{ }/180.**

[This can be remembered as ** t_{C }= (5/9)(t_{F }– 32)**]

_{ }

Here *t*_{C} =* **t* and *t*_{F} = 2*t* so that *t/*100 = (2*t*– 32)_{ }/180.

Or, 9*t =* 10*t*– 160 from which *t = *160.

Therefore, the _{ }

**(3)** A copper rod and a steel rod are to have lengths *L*c* *and* L*_{s} such that the *difference* between their lengths is the *same* at all ambient *α*_{c} and* **α*_{s} the lengths are related to the coefficients of linear expansion as

(a) *L*c/*L*_{s} = *α*_{c}/*α*_{s}

(b) *L*c – *L*_{s} = *α*_{c }– *α*_{s}

(c) *L*c/*L*_{s} = *α*_{s}/*α*_{c}

(d) *L*c/*L*_{s} = (*α*_{c}/*α*_{s})^{1/2}

(e) *L*c/*L*_{s} = (*α*_{c}/*α*_{s})^{–1/2}

The changes in length must be equal on heating (or cooling) so that we have

*L*c* **α*_{c }*∆T *=* L*s* **α*_{s }*∆T*

This gives *L*c/*L*_{s} = *α*_{s}/*α*_{c}

**(4) **Four identical brass rods AB, BC, CD and DA are joined to form a square (fig). If the diagonally opposite corners A and C are kept in melting ice and boiling water respectively what is the

(a) 50º C

(b) 100º C

(c) 25º C

(d) 75º C

(e) 0º C

The corners B and D will be at the *same* *difference* will be zero.

**(5) **Thermal conductivity of metal A is three times the thermal conductivity of metal B. Two rods of the same dimensions made of metals A and B are joined end to end (fig.). The free end of A is maintained at 0º C and that of of B is maintained at 100º C. The

(a) 25º C

(b) 33.3º C

(c) 66.6º C

(d) 75º C

(e) 90º C

In the steady state the quantity of heat flowing through B per second will be equal to that flowing through A per second. If the *t*º C we have

*k*A(100 – *t*)/*L * = 3*k*A(*t* – 0)/*L * where A is the area of cross section, *L* is the length, *k* is the thermal conductivity of B and therefore 3*k *is the thermal conductivity of A.

Therefore 100 – *t * = 3*t *so that *t = *25º C

[Since the rods are of identical dimensions and the thermal conductivity of A is three times that of B, you can easily argue that the

** (6) **A cup of coffee cools from 65º C to 63º C in one minute in a room of

(a) 2 min.

(b) 2.4 min.

(c) 3.8 min.

(d) 4.5 min

(e) 5.3 min

Since the rate of cooling is directly proportional to the excess of

*S*(65 – 63)/1 α (64 – 24) and

*S*(58 – 50)/*t* α (54 – 24)

where *S* is the *heat capacity* (thermal capacity) of the cup of coffee (which is the heat required to raise the temperature through 1K) and *t* is the time required to cool from 58º C to 50º C. Note that we have used the *average*

On dividing, 2*t/*8* =* 40/30 from which *t = *5.3 min (nearly)

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