Question

JEE ADVANCED PHYSICS QUESTION SEPTEMBER 2020

Answer 1

In the balance condition, the values of the resistance of the four arms of a Wheatstone's bridge are shown in figure. The resistance R₃ has temperature coefficient 0.0004/°C .

To find : if the temperature of R₃ is increased by 100°C, the voltage developed between S and T will be ....

solution : using formula, R = R₀[1 + α∆T]

initial resistance of R₃ = 300 Ω

coefficient of Temperature , α = 0.0004/°C

and temperature change, ∆T = 100°C

so, R₃ = 300[1 + 4/10000 × 100] = 312 Ω

now,

let i₁ is passing through left loop and i₂ is passing through right loop as shown in figure.

for first loop,

i₁ = 50/(60 + 312) = 50/372

for 2nd loop,

i₂ = 50/(100 + 500) = 50/600

so, voltage between S and T = Vs - VT = 312i₁ - 500i₂

= 312 × 50/372 - 500 × 50/600

= 41.94 - 41.67

= 0.27 volts

Therefore the voltage between s and T is 0.27 volts.

Answer 2

Transpiration is the process of water movement through a plant and its evaporation from aerial parts, such as leaves, stems and flowers. Water is necessary for plants but only a small amount of water taken up by the roots is used for growth and metabolism. The remaining 97–99.5% is lost by transpiration and guttation.