## An electric dipole of moment p = (-i-3j+2k) × 10^-29 C m is at the origin (0,0,0) …..

Q: An electric dipole of moment $\displaystyle \vec{p} = (-\hat{i}-3\hat{j} + 2\hat{k}) \times 10^{-29} C m$ is at the origin (0,0,0) . the electric field due to this dipole at $\displaystyle \vec{r} = \hat{i} + 3 \hat{j} + 5 \hat{k}$ (note that $\vec{r}.\vec{p} = 0$) is parallel to

(a) $\displaystyle (\hat{i} – 3 \hat{j}-2\hat{k})$

(b) $\displaystyle (-\hat{i} – 3 \hat{j} + 2\hat{k})$

(c) $\displaystyle (\hat{i} + 3 \hat{j}-2\hat{k})$

(d) $\displaystyle (- \hat{i} + 3 \hat{j}- 2\hat{k})$

Click to See Solution :
Ans: (c)

Sol: $\displaystyle \vec{p} = (-\hat{i}-3\hat{j} + 2\hat{k}) \times 10^{-29} C m$

$\displaystyle \vec{r} = \hat{i} + 3 \hat{j} + 5 \hat{k}$

As , $\displaystyle \vec{r}.\vec{p} = 0$ , It means $\vec{r}$ is perpendicular to $\vec{p}$ Therefore point lies on equilateral plane . Thus electric field is anti-parallel to dipole moment or $\vec{E}|| (-\vec{p})$

$\displaystyle \vec{E}|| (-(-\hat{i}-3\hat{j} + 2\hat{k}))$

$\displaystyle \vec{E}|| (\hat{i} + 3\hat{j} – 2\hat{k})$

## Two infinite planes each with uniform surface charge density +σ are kept in such a way that ….

Q: Two infinite planes each with uniform surface charge density +σ are kept in such a way that the angle between them is 30° . The electric field in the region shown between them is given by

(a) $\displaystyle \frac{\sigma}{\epsilon_0}[(1+\frac{\sqrt{3}}{2})\hat{y} + \frac{\hat{x}}{2}]$

(b) $\displaystyle \frac{\sigma}{2\epsilon_0}[(1+\sqrt{3})\hat{y} + \frac{\hat{x}}{2}]$

(c) $\displaystyle \frac{\sigma}{2\epsilon_0}[(1+\sqrt{3})\hat{y} – \frac{\hat{x}}{2}]$

(d) $\displaystyle \frac{\sigma}{2\epsilon_0}[(1-\frac{\sqrt{3}}{2})\hat{y} – \frac{\hat{x}}{2}]$

Click to See Solution :
Ans: (d)
Sol: The electric field due to charge sheet of charge density +σ is $E = \frac{\sigma}{2\epsilon_0}$

Electric field at point P is

$\displaystyle \vec{E} = \frac{\sigma}{2 \epsilon_0}cos60^0 (-\hat{x}) + \frac{\sigma}{2 \epsilon_0}sin60^0 (-\hat{y}) + \frac{\sigma}{2 \epsilon_0} \hat{y}$

$\displaystyle \vec{E} = \frac{\sigma}{2\epsilon_0}[(1-\frac{\sqrt{3}}{2})\hat{y} – \frac{\hat{x}}{2}]$

## If minimum possible work is done by a refrigerator in converting 100 gm of water at 0°C to ice ….

Q: If minimum possible work is done by a refrigerator in converting 100 gm of water at 0°C to ice , how much heat (in calorie) is released to the surroundings at temperature 27 °C (Latent heat of ice = 80 cal/gram) to the nearest integer ?

Click to See Solution :
Ans: (8791)

Sol: T1 = 27 °C = 300 K , Lice = 80 cal/g

Heat energy extracted from water , Q2 = m L

Q2 = 100 x 80 = 8000 cal

T2 = 0 °C = 273 K

$\displaystyle \frac{Q_1}{Q_2} = \frac{T_1}{T_2}$

$\displaystyle Q_1 = Q_2 \times \frac{T_1}{T_2}$

$\displaystyle Q_1 = 8000 \times \frac{300}{273}$

= 8791 cal

## A bakelite beaker has volume capacity of 500 cc at 30 °C .When it is partially filled with volume Vm at 30°C of mercury ….

Q: A bakelite beaker has volume capacity of 500 cc at 30 °C .When it is partially filled with volume Vm at 30°C of mercury . , it is found that the unfilled volume of the beaker remains constant as the temperature is varied . If γbeaker = 6 × 10-6/°C and γmercury = 1.5 × 10-4/°C , where γ is the coefficient of volume expansion , then Vm in cc is close to ……

Click to See Solution :
Ans: (20)

Sol: Given V = 500 cc , T = 30 °C

γbeaker = 6 × 10-6/°C

and γmercury = 1.5 × 10-4/°C

V – Vm = V’ – Vm

Now , V’ = V(1 + γb ΔT)

and , Vm‘ = Vm(1 + γm ΔT)

V – Vm = V + V γb ΔT – Vm – Vm γm ΔT)

$\displaystyle V_m = V \frac{\gamma_b}{\gamma_m}$

$\displaystyle V_m = 500 \times \frac{6 \times 10^{-6}}{1.5 \times 10^{-4}}$

Vm = 20 cc

## A non-isotropic solid metal cube has coefficients of linear expansion as 5 × 10-5 °/C  along the x -axis ….

Q: A non-isotropic solid metal cube has coefficients of linear expansion as 5 × 10-5 °/C  along the x -axis and 5 × 10-6 °/C  along the y -axis and z -axis . If the coefficient of volume expansion of solid is C × 10-5 °/C then the value of C is ….

Click to See Solution :
Ans: (60)

Sol: αlx = 5 × 10-5 °/C

αly = αlz = 5 × 10-6 °/C

αv = αlx + αly + αlz

αv = αlx + 2 αly

αv = 5 × 10-5 + 2 × 5 × 10-6

= 6 × 10-5 °/C

= 60 × 10-6 °/C

C = 60