Bihar Board Physics Class 12 Question Papers 2018 with Solutions - Free PDF Download.
FAQs on Bihar Board (BSEB) Question Paper for Class 12 Physics 2018.
1. Find out the total resistance differently when four resistors each have 12Ω are connected in parallel and three 3Ω type resistors are linked in series.
For series connection, RP = 3 + 3 + 3 = 9Ω
For parallel connection, 1/RP = 1/r + 1/R + …. + n = n / R (n = no. of resistors)
So, RP = R/n = 12/4 = 3Ω
2. A capacitor connected to the 200V power with parallel the parallel plates of area 6cm2 kept at a separation of 2mm. If the air between the plates acts as a dielectric medium, calculate the capacitance of the parallel plate capacitor.
Here, Area, A = 6cm2 = 6 * 10-4 m3
Separation, d = 2 mm
= 2 * 10-3 m
Then,
Capacitance, C = (A × E0)/(d) = (6 x10-4 × 6×10-4) /(2 × 10-3)= 2.655 x 10-12 f
3. What remains constant in a nuclear reaction?
These are the factors that remain conserved in nuclear reactions:
Charge
Energy
momentum
Angular momentum
4. What helps to incline the value of the specific resistance of a conductor?
The rise in temperature helps to increase the value of the specific resistance of a conductor. The reason is the specific resistance of a conductor is directly proportional to the temperature of the conductor.