NEETshala
Log in Get Started Free

NEET Practice Questions (MCQs) with Answers & Solutions

Practice free NEET NEET multiple-choice questions online with instant answers and detailed explanations. No login required.

All Physics Chemistry Botany Zoology
Register free for language, difficulty & keyword filters
● ● ● ● ●

How would you arrange 48 cells each of e.m.f 2V and inteanal resistance $1.5 \Omega $ so as to pass maximum current through the external resistance of $2 \Omega$ ?

You've reached today's free limit of 20 questions. Log in to keep practising for free.
Explanation

[{No of rows}/{no of cells in each row}] = (m/n) = (r/R) where r = internal resistance of each cell

R = external resistance

∴ (m/n) = {1.5/2} = (3/4)

also total number of cells = 48 m × n = 48

∴ (3/4) n × n = 48 n2 = {(48 × 4)/3} n = 8

∴ m = (48/8) = 6

∴ 6 rows with 8 cell in each of row

● ● ● ● ●

How many dry cells, each of emf 1.5V and internal resistance $0.5 \Omega $, much be joined in series with a resistor of $ 20 \Omega $ to give a current of 0.6A in the circuit ?

You've reached today's free limit of 20 questions. Log in to keep practising for free.
Explanation

$ l = { n \varepsilon \over R +nr } \Leftrightarrow 0.6 = { n \times 1.5 \over 20 \times 0.5 n } \Leftrightarrow 12 + 0.3 n = 1.5 n \therefore 12 = 1.2 n $ $ \therefore n = 10 $cells to be connected in series

● ● ● ● ●

Two electric bulbs whose resistances are in the ratio of 1:2 are connected in parallel to a constant voltage source the power dissipated in them have the ratio.

You've reached today's free limit of 20 questions. Log in to keep practising for free.
Explanation

$ {P_1/P_2 } = {V^2 / R_1 \over V^2 / R_2 } = {R_2 \over R_1} = 2 /1 $

● ● ● ● ●

An electric kettle has two coils. when onc of them is switched on, the water in the kettle boils in 6 minutes. When the other coil is switched on, the water boils in 3 minutes If the two coils are connected in series the time taken to boil water in the kettle is:

You've reached today's free limit of 20 questions. Log in to keep practising for free.
Explanation

$H_1 = { V^2 \over R_1 } t_1$ $ R_1 = { V^2 \over H_1} t_1 and R_2 = { V^2 \over H_2 } t_2$ $ { V^2 \over H_1 } = { V^2 \over H_2 } = const \therefore R\, \alpha \,t $ $ Now R_s = R_1 + R_2 $ t = 6 + 3 = 9 Minutes

● ● ● ● ●

An electric kettle has two coils when one of these is switched on. the water in the kettle boils in 6 minutes. When the other coil is switched on, boils in 3 minutes If the two coils are connected in parallel, the time taken to boil water in the kettle is

You've reached today's free limit of 20 questions. Log in to keep practising for free.
Explanation

$R_p = { R_1 R_2 \over R_1 + R_2 } $ $ t = {6 \times 3 \over 6 +3} = 2 minutes $

● ● ● ● ●

The potential difference between the terminals of a battery is 10V and internal resistance $1 \Omega $ drops to 8V when connected across an external resistor find the resistance of the external resistor.

You've reached today's free limit of 20 questions. Log in to keep practising for free.
Explanation

$ V = IR \,in I = { E \over R +r } $ $ \therefore ={ ER \over R + r } $

● ● ● ● ●

A heater boils 1kg of water in time $t_1$ and another heater boils the same water in time $t_2$ If both are connected in series, the combination will boil the same water in time.

You've reached today's free limit of 20 questions. Log in to keep practising for free.
● ● ● ● ●

At what temperature will the resistance of a copper wire be three times its value at 0°C ? (Given: temperature coefficient of resistance for copper = $4 \times 10^{-3} C^{-1}$ )

You've reached today's free limit of 20 questions. Log in to keep practising for free.
Explanation

$R_\theta = R_\theta 0 [ 1 + \alpha ( \theta _2 - \theta _1 ) ] $ $ 3 = 1 + ( 4 \times 10 ^ {-3} ) T$ $ \therefore T = { 2 \over 4 \times 10 ^ {-3}} = 500 C $

● ● ● ● ●

The resistance of a copper coil is $4.64 \Omega $ at 40 °C and $5.6 \Omega$ at 100 °C Its resistcnce at 0° C will be

You've reached today's free limit of 20 questions. Log in to keep practising for free.
Explanation

$R_{40} [1 + \alpha (\theta_2 - \theta_1)]$ $ 5.6 = 4.64[1 + a(100 -40)] =4.64+278.4 a$ $a ={5.6 -4.64 \over 278.4} = 0.0035 C^{-1}$ $ \Rightarrow _{100} = R_0(1+ \alpha T_2) = R_0 [1+(0.0035 \times 100)]$ $ 5.6 = R_0 \times 1.35$ $Ro = 4 \Omega $

● ● ● ● ●

There are n resistors having equal value of resistance r. First they are connected in such a way that the possible minimum value of resistance is obtained. Then they are connected in such a way that possible maximum value of resistance is obtained the ratio of minimum and maximum values of resistances obtained in these way is....

You've reached today's free limit of 20 questions. Log in to keep practising for free.
Explanation

In parallal connection R minimum = r/n In Series connection R maximum= nr $ { R min \over R max } = { r \over n(n-1) } = {1 \over n^2}$

Ready to ace NEET?

Free access · No credit card required

Browse All Questions Create Free Account

Frequently Asked Questions

Yes. You can attempt every NEET question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.