As per analysis for previous years, it has been observed that students preparing for NEET find Physics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Biology due to their medical background. Furthermore, sections such as Physics are dominantly based on theories, laws, numerical in comparison to a section of Biology which is more of fact-based, life sciences, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced.
.

**Q1.**In the given figure, battery E is balanced on 55 cm length of potentiometer wire but when a resistance of 10 Ω is connected in parallel with the battery then it balances on 50 cm length of the potentiometer wire then internal resistance r of the battery is?

Solution

r=((l

r=((l

_{1}-l_{2})/l_{2})×R^{'}⇒r=((55-50)/50)×10=1Ω**Q2.**Which statement is true? (i) Kirchoff’s law is equally applicable to both AC and DC. (ii) Semiconductors have a positive temperature coefficient of resistance. (iii) Meter bridge is greater sensitive when the resistance of all four arms of the bridge is of the same order. (iv) The emf of a cell depends upon the size and area of electrodes.

Solution

In a potentiometer there is no current drawn from the cell whose emf is to be measured whereas a voltmeter always draws some current from the cell. Hence, the emf of a cell can be measured accurately using a potentiometer.

In a potentiometer there is no current drawn from the cell whose emf is to be measured whereas a voltmeter always draws some current from the cell. Hence, the emf of a cell can be measured accurately using a potentiometer.

**Q3.**An electric iron draws 5 amp, a TV set draws 3 amp and refrigerator graws 2 amp from a 220 volt main line. The three appliances are connected in parallel. If all the three are operating at the same time, the fuse used may be of?

Solution

Current capacity of a fuse wire should be slightly greater than the total rated load current

Current capacity of a fuse wire should be slightly greater than the total rated load current

**Q4.**The ratio of the amounts of heat developed in the four arms of a balanced Wheatstone bridge, when the arms have resistance P=100 Ω; Q=10 Ω;R=300 Ω and S=30 Ω respectively is?

Solution

Let I be the total current passing through balanced wheat stone bridge. Current through arms of resistances P and Q in series is I

Let I be the total current passing through balanced wheat stone bridge. Current through arms of resistances P and Q in series is I

_{1}=(I×300)/(330+110)=3/4 I and current through arms of Resistances R and S in series is I_{2}=(I×110)/(330+110)=1/4 I ∴ Ratio of heat developed per sec H_{P}∶H_{Q}∶H_{R}∶H_{S}=(3/4 I)^{2}×100∶(3/4 I)^{2}×10∶(1/4 I)^{2}×300∶(1/4 I)^{2}×30 =30∶3∶10∶1.**Q5.**Length of a hollow tube is 5m, it’s outer diameter is 10 cm and thickness of it’s wall is 5 mm. If resistivity of the material of the tube is 1.7×10

^{-8}Ω×m then resistance of tube will be?

**Q6.**A milliammeter of range 0-30mA has internal resistance of 20 Ω. The resistance to be connected in series to convert it into a voltmeter of maximum reading 3V is?

Solution

R= V/I

R= V/I

_{g}-G=3/(30×10^{-3})-20=10^{2}-20=80Ω**Q7.**A lamp having tungsten filament consumes 50 W. Assume the temperature coefficient of resistance for tungsten is 4.5×10

^{-3}℃

^{-1}and temperature of the surrounding is 20℃. When the lamp burns, the temperature of its filament becomes 2500℃ , then the power consumed at the moment switch is on, is?

Solution

Let R

Let R

_{0}= resistance of filament at room temperature R_{t}= resistance of filament at 2500℃ Similarly powers, P_{0}and P_{t}. Here, voltage remains the same. P_{0}=V^{2}/R_{0}or R_{0}=V^{2}/P_{0}, R^{t}=V^{2}/P_{t}also R_{t}=R_{0}[1+α(2500-20) ] and P_{0}=P_{t}[1+α(2500-20)] =50[1+4.5×10^{-3}(2500-20)] = 608W**Q8.**A heater coil cut into two equal parts and one part is connected with heater. Now heat generated in heater will be?

Solution

Half

Half

**Q9.**A long straight wire of a circular cross section (radius a) carries a steady current I and the current I is uniformly distributed across this cross-section. Which of the following plots represents the variation of magnitude of magnetic field B with distance r from the centre of the wire?

Solution

Magnetic field due to a long straight wire of radius a carrying current I at a point distant r from the centre of the wire is given as follows B=(μ

Magnetic field due to a long straight wire of radius a carrying current I at a point distant r from the centre of the wire is given as follows B=(μ

_{0}Ir)/(2πa^{2}) for r less than a B=(μ_{0}I)/2πr for r=a B=(μ_{0}I)/2πr for r>a The variation of magnetic field B with distance r from the centre of wire is shown in the figure**Q10.**The resistance of a discharge tube is?

Solution

This is because of secondary ionisation which is possible in the gas filled in it

This is because of secondary ionisation which is possible in the gas filled in it