Moving Charges And Magnetism

Step 1: Convert energy per fission from MeV to joules.
Step 2: Power is energy released per second. Given power:
Step 3: Fission rate required:

Step 1: From the circuit, the supply voltage is: Collector resistance:
Step 2: Given: Since the emitter is grounded, the collector voltage is:
Step 3: Voltage drop across the collector resistance:
Step 4: Collector current:
Step 5: Using the relation :
Step 6: The closest value among the given options is:

Step 1: Recall Bohr model dependences. For a hydrogen atom, the Bohr radius and energy depend on the reduced mass of the electron–proton system. For ordinary hydrogen:
Step 2: Effect of doubling the electron mass. If the electron mass is doubled:
Step 3: Find the new radius of the first orbit. Since:
Step 4: Find the new ground-state energy. Since: Original ground-state energy: With doubled mass:
Hence,

Step 1: Identify the given quantities. Angle between magnet and magnetic meridian: Apparent angle of dip:
Step 2: Recall the relation between real dip and apparent dip. When the magnet is not in the magnetic meridian, the relation is: where = real angle of dip.
Step 3: Substitute the given values.
Step 4: Solve for the real angle of dip.
Hence, the real angle of dip is .

Step 1: Recall the formula for electric potential of a charged sphere. For a conducting sphere: where is a constant and is the radius of the sphere. Step 2: Compare the given spheres. The spheres are identical, so their radii are the same. Hence, electric potential depends only on the charge. Step 3: Compare the charges. Since: corresponds to a higher (less negative) potential. Step 4: Final conclusion. The sphere carrying charge is at a higher potential.

Step 1: Examine statement (1). A discharge lamp works by passing electric current through a gas at very low pressure inside a discharge tube (e.g., neon lamp, sodium vapour lamp). Hence, statement (1) is true. Step 2: Examine statement (2). The colour of light emitted by a discharge lamp depends on the type of gas used. For example:

Neon gas → red light
Sodium vapour → yellow light
Mercury vapour → bluish light
Therefore, white light is not always emitted. Statement (2) is false. Step 3: Final conclusion. Only statement (1) is true.