CBSE Questions for Class 10 Physics Electricity Quiz 10 - MCQExams.com

$$(1) \& (1)$$ points same potential $$\& (2) \& (2)$$ points are same potential. So both the corner Resistance shorted.

Most above with has no resistance. So all are shorted.

$$I=\dfrac{2}{1}=2A$$

• Two resistors when connected in parallel always have equivalent resistance less than or equal to the resistance of the smaller resistor.  $$\\ \dfrac 1R_{eq} = \dfrac 1R_1 + \dfrac 1R_2$$
• Since the smaller resistance is $$6 \Omega$$ ,the resistance of $$R_1$$ has to be greater than or equal to $$6 \Omega$$.
  

Given that,

The dimension of the block is 1cm, 3cm and 3cm.

The relation of the resistance of a conductor is

$$R=\rho \dfrac{l}{A}....(I)$$

Where,

$$\rho $$ = resistivity of the material of the conductor

$$A$$= area of the cross section of the conductor

$$l$$= length of the conductor

Now, the resistance is maximum

$$ {{R}_{\max }}=\rho \times \dfrac{3}{1\times 2} $$

$$ {{R}_{\max }}=\dfrac{3\rho }{2}...(II) $$

Now, the resistance is minimum

$$ {{R}_{\min }}=\rho \dfrac{1}{2\times 3} $$

$$ {{R}_{\min }}=\dfrac{\rho }{6}...(III) $$

Now, from equation (II) and (III)

The ratio of maximum resistance and minimum resistance is

$$ \dfrac{{{R}_{\max }}}{{{R}_{\min }}}=\dfrac{\frac{3\rho }{2}}{\dfrac{\rho }{6}} $$

$$ \dfrac{{{R}_{\max }}}{{{R}_{\min }}}=\dfrac{3}{2}\times \dfrac{6}{1} $$

$$ \dfrac{{{R}_{\max }}}{{{R}_{\min }}}=\dfrac{9}{1} $$

$$ {{R}_{\max }}:{{R}_{\min }}=9:1 $$

Hence, the ratio is $$9:1$$

Resistance in wire, $$R=\rho \dfrac{L}{A}=\dfrac{\rho L}{\pi {{r}^{2}}}$$

  $$ {{R}_{1}}=\dfrac{\rho {{L}_{1}}}{\pi r_{1}^{2}}=\dfrac{\rho {{L}_{1}}}{\pi {{(2r)}^{2}}}=\dfrac{1}{4}\dfrac{\rho {{L}_{1}}}{\pi {{r}^{2}}}=34\,\Omega  $$

 $$ \Rightarrow \dfrac{\rho {{L}_{1}}}{\pi {{r}^{2}}}=34\times 4=136\ \Omega \ ......\ (1) $$

$$L_2=L_1/2$$      $$r_2=r$$

 $$ {{R}_{2}}=\rho \dfrac{{{L}_{2}}}{{{A}_{2}}}=\dfrac{\rho \dfrac{{{L}_{1}}}{2}}{\pi r_{2}^{2}}=\dfrac{1}{2}\dfrac{\rho {{L}_{1}}}{\pi {{r}^{2}}}=\dfrac{1}{2}\times 136=\ 68\,\Omega  $$

Resistance of B is $$68\,\Omega .$$ 

$$\large\textbf{Step -1: Calculate current For Electric bulbs}$$

Given,

Power of heater $$P_h= 1500W,$$

Power of bulbs,$$P_b = 100W,$$

We know,

Power,$$ P = VI$$

The current drawn by the heater,

Or $$Electrical\;Current,\;I\; = \dfrac{P_h}{V} = \dfrac{{1500}}{{220}} = 6.81\;A$$

The mains power supply of a house is through a 20 A fuse

Maximum Electric current, which can pass through bulbs $$= 20- 6.81 =13.19 A$$

$$\large\textbf{Step -2:Calculate number of bulbs connected in parallel}$$

Let number of bulb $$= n$$

and Current withdraw by a single bulb$$, I_b$$

 $$Electrical\;Current,\;{I_b}\; = \dfrac{P_b}{V} = \dfrac{{100}}{{220}} = 0.45\;A$$

Maximum number of bulbs, $$n= 13.19/0.45 = 29$$

Hence maximum,$$ 29 $$ bulbs can be connected in parallel with heater.

R₁  = 24 ± 0.5

R₂ = 8 ± 0.3

In series R₁ + R₂  = 32 ± 0.8