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.
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**Q1.**When a piece of aluminium wire of finite length is drawn through a series of dies to reduce its diameter to half its original value, its resistance will become?

Solution

In stretching of wire R∝1/d

In stretching of wire R∝1/d

^{4}, where d= Diameter of wire
Solution

The current in the circuit =8/(5+1)=4/3 Now V

The current in the circuit =8/(5+1)=4/3 Now V

_{C}-V_{E}=4/3×1⇒V_{E}=-4/3 V**Q3.**An electric fan and a heater are marked as 100 watt,220 volt and 1000 watt,220 volt respectively. The resistance of the heater is?

Solution

For constant voltage, we know that P∝1/R So higher the power, lower will be the resistance

For constant voltage, we know that P∝1/R So higher the power, lower will be the resistance

**Q4.**In a thermocouple, the neutral temperature is 270℃ and the temperature of inversion is 525℃. The temperature of cold junction would be?

Solution

15℃

15℃

**Q5.**5 ampere of current is passed through a metallic conductor. The charge flowing in one minute in coulomb will be?

Solution

Charge = Current × Time =5×60=300 C

Charge = Current × Time =5×60=300 C

**Q6.**When the resistance of 9 Ω is connected at the ends of a battery, its potential difference decreases from 40 volt to 30 volt. The internal resistance of the battery is?

Solution

The internal resistance of battery is given by r=(E/V-1)R=(40/30-1)×9=(9×10)/30=3Ω

The internal resistance of battery is given by r=(E/V-1)R=(40/30-1)×9=(9×10)/30=3Ω

**Q7.**A wire has a resistance of 6Ω. It is cut into two parts and both half values are connected in parallel. The new resistance is?

Solution

Resistance of original wire is R=ρ l/A ρ,being the specific resistance of wire. When the wire is cut in two equal halves then resistance becomes R

Resistance of original wire is R=ρ l/A ρ,being the specific resistance of wire. When the wire is cut in two equal halves then resistance becomes R

^{'}=(ρl/2)/A=R/2 Thus, the net resistance of parallel combination of two halves is given by R_{net}=(R^{'}×R^{'})/(R^{'}+R^{'}) = R^{'}/2=R/(2×2)=6/4=1.5Ω**Q8.**A conductor wire having 10

^{29}free electrons/m

^{3}carries a current of 20A. If the cross-section of the wire is 1mm

^{2}, then the drift velocity of electrons will be?

Solution

v

v

_{d}=I/nAe=20/(10^{29}×10^{-6}×1.6×10^{-19})=1.25×10^{-3}m/s**Q9.**Above neutral temperature, thermo e.m.f. in a thermocouple?

Solution

Decreases with rise in temperature

Decreases with rise in temperature