ICSE Physics

Sponsor Area

Question
ICSEENIPH12030210

An electric dipole of dipole moment p is placed in a uniform electric field with its axis inclined to the field. Write an expression for the torque straight tau experienced by the dipole in vector form. Show diagrammatically how the dipole should be kept in the electric field so that torque acting on it is:

i) maximum

ii) zero

Solution

Expression for torque in the vector form is:
straight tau space equals space straight p with rightwards harpoon with barb upwards on top space straight x space straight E with rightwards harpoon with barb upwards on top ;
where,
p is the dipole moment, and
E is the electric field.
i) When torque acting is maximum, dipole should be placed perpendicular to the electric field.
          
ii) When torque acting is zero, dipole is placed parallel to the magnetic field.
                          

Sponsor Area

Question
ICSEENIPH12030211

You are provided with 8 μF capacitors. Show with the help of a diagram how you will arrange minimum number of them to get a resultant capacitance of 20 μF.

Solution

                      
Any two capacitors are connected in series and the othr two capacitors are connected in parallel.

Each capacitor is 8 straight mu space straight F

Question
ICSEENIPH12030212

i) Define temperature coefficient of resistance of the material of a conductor.

ii) When the cold junction of a thermocouple is maintained at 0o C, the thermo emf 'e', generated by this thermocouple is given by the relation,

straight e space equals space left square bracket 16.8 space straight theta space plus space 1 half left parenthesis negative 0.048 space straight theta squared right square bracket space straight x space 10 to the power of negative 6 end exponent,
where straight theta is the temperature of the hot junction in oC. Find the neutral temperature of the thermocouple.

Solution

i) Temperature coefficient of resistance is defined as fractional increase in resistance per degree rise in temperature.
ii)
straight e space equals space left square bracket 16.8 space plus 1 half space left parenthesis negative 0.048 right parenthesis space straight theta squared right square bracket space straight x space 10 to the power of negative 6 end exponent
de over dθ space equals space left square bracket 16.80 space minus space 1 half straight x 0.048 space straight x space 2 straight theta right square bracket space straight x space 10 to the power of negative 6 end exponent

de over dθ space equals space 0

rightwards double arrow space straight theta space equals space 350 to the power of straight o space straight C

Question
ICSEENIPH12030213

Draw a labelled circuit diagram of a potentiometer to compare emf of two cells. Write the working formula. 

Solution

The potentiometer to compare EMFs of two cells is as shown below:

Working formula is given by, 
E1 : E2 = l1 : l2

Question
ICSEENIPH12030214

b) How much resistance should be connected to 15-ohm resistor shown in the circuit in figure below so that the points M and N are at the same potential.

Solution

For points M and N to be at the same potential, the resistance that should be connected to 15 ohm is:

space space space 3 over straight R space equals 30 over 60

rightwards double arrow space R space equals space 6 space capital omega

space space space space 1 over 15 space plus space 1 over x space equals space 1 over 6

rightwards double arrow space fraction numerator 15 space x over denominator 15 space plus space x end fraction space equals space 6

rightwards double arrow space x space equals space 10 capital omega