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Day 155: JEE Mains Physics

5 questions. Try each one first, then open its answer.

  1. Question 1

    Moving Charges and Magnetism. Asked once: 30 Jan 2023 (same type asked 6 times) See all 6 of this type

    91
    A current carrying rectangular loop PQRSPQRS is made of uniform wire. The length PR=QS=5 cmPR = QS = 5\text{ }cm and PQ=RS=100 cmPQ = RS = 100\text{ }cm. If ammeter current reading changes from II to 2I2I, the ratio of magnetic forces per unit length on the wire PQPQ due to wire RSRS in the two cases respectively (fPQI:fPQ2I)\left( f_{PQ}^{I}:f_{PQ}^{2I} \right) is:

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  2. Question 2

    System of Particles and Rotational Motion. Asked once: 5 Apr 2024 (same type asked 6 times) See all 6 of this type

    11
    Ratio of radius of gyration of a hollow sphere to that of a solid cylinder of equal mass, for moment of Inertia about their diameter axis ABAB as shown in figure is 8x\sqrt{\frac{8}{x}}. The value of xx is:

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  3. Question 3

    Current Electricity. Asked once: 22 Jan 2026 S1 (same type asked 6 times) See all 6 of this type

    6
    A meter bridge with two resistances R1R_{1} and R2R_{2} as shown in figure was balanced (null point) at 40 cm from the point P. The null point changed to 50 cm from the point P, when 16Ω16\Omega resistance is connected in parallel to R2R_{2}. The values of resistances R1R_{1} and R2R_{2} are ____\_\_\_\_ .

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  4. Question 4

    Moving Charges and Magnetism. Asked once: 23 Jan 2025 (same type asked 6 times) See all 6 of this type

    87
    A galvanometer having a coil of resistance 30Ω30\Omega need 20 mA of current for full-scale deflection. If a maximum current of 3 A is to be measured using this galvanometer, the resistance of the shunt to be added to the galvanometer should be 30XΩ\frac{30}{X}\Omega, where X is

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  5. Question 5

    Moving Charges and Magnetism. Asked once: 24 Jan 2025 (same type asked 6 times) See all 6 of this type

    24
    A current of 5 A exists in a square loop of side 12 m\frac{1}{\sqrt{2}}\text{ }m. Then the magnitude of the magnetic field BB at the centre of the square loop will be p×10−6 Tp \times10^{- 6}\text{ }T. where, value of p is _____\_\_\_\_\_ [Take μ0\left\lbrack Take\ \mu_{0} \right. =4π×10−7 T mA−1= 4\pi\times10^{- 7}\text{ }T{\text{ }mA}^{- 1} ].