Home
Class 12
PHYSICS
In a PO box experiment a student obserfe...

In a PO box experiment a student obserfes that he does not get the null point for any of the ratios. But when the ratio is `1000:10` he finds that the deflectioin is to the right by 1 division for `596Omega` in the rheostat arm and 0.2 direction to the left when it is `597Omega`. Compute the correct value of the unkown resistance from the observations of the student.

Text Solution

Verified by Experts

The correct Answer is:
5.9683 `Omega`
Promotional Banner

Topper's Solved these Questions

  • WHEATSTONE BRIDGE, METER BRIDGE

    NN GHOSH|Exercise Exercise|10 Videos

  • TERRESTRIAL MAGNETISM

    NN GHOSH|Exercise All Questions|7 Videos

Similar Questions

Explore conceptually related problems

In a PO box experiment it is found that for the 1000:10 ratio, the deflection is to the left by 2 divisions when R is 437Omega and to the right by 1.3 cm division when R is 436Omega . Compute the correct value of the unknown resistance.

In A PO box experiment it is found that for the 100:10 ratio deflection in the galvanometer is to the left by 0.7 division of the scale when resistance I the rheostat are is 978Omega and to the right by 1.2 division when is 979Omega . Calculate the unknown resistance.

In a meter bridge experiment the resistance to be meausred in connected int the right gap and a known resistance in the left gap has value of 50+-0.2Omega when the null points is judged to be at 60+-2.0 cm. The students notes is judged to be of the bridge wire are not at 0.0 cn and 100.0 cm of the scale makes ends. The value of the unknown resistance should be expressed as

Experiment set up of a meter bridge .When the two unknown resistance X and Y are intserted, the null point D is obtained 40 cm from the end A .When a resistance of 30 Omega is connected in series with X the mill point shift by 10 cm .Find the position of the point when the 30 Omega resistance is connected in series with X, the null point shifts by 10 cm. Find the position of the null point when the 30 Omega resistance is connected in series with resistance Y instead of X . Determine the values of resistances X and Y

An unknown resistance is connected in a left gap of a metre bridge . The balance point is obtained when a resistance of 10 Omega is taken out from the resistance box in the right gap. On increasing the resistance from the resistance box by 12.5 Omega the balance point shifts by 20 cm. find the unknown resistance.

In a meter bridge experiment the resistance of resistance box is 16 Omega which is inserted in right gap The null point is obtained at 36cm from the left end the least count of meter scale is lmm What is the value of unknown resistence ?( Error = L.C or L.C /2 )

Direction : Resistive force proportional to object velocity At low speeds, the resistive force acting on an object that is moving a viscous medium is effectively modeleld as being proportional to the object velocity. The mathematical representation of the resistive force can be expressed as R = -bv Where v is the velocity of the object and b is a positive constant that depends onthe properties of the medium and on the shape and dimensions of the object. The negative sign represents the fact that the resistance froce is opposite to the velocity. Consider a sphere of mass m released frm rest in a liquid. Assuming that the only forces acting on the spheres are the resistive froce R and the weight mg, we can describe its motion using Newton's second law. though the buoyant force is also acting on the submerged object the force is constant and effect of this force be modeled by changing the apparent weight of the sphere by a constant froce, so we can ignore it here. Thus mg - bv = m (dv)/(dt) rArr (dv)/(dt) = g - (b)/(m) v Solving the equation v = (mg)/(b) (1- e^(-bt//m)) where e=2.71 is the base of the natural logarithm The acceleration becomes zero when the increasing resistive force eventually the weight. At this point, the object reaches its terminals speed v_(1) and then on it continues to move with zero acceleration mg - b_(T) =0 rArr m_(T) = (mg)/(b) Hence v = v_(T) (1-e^((vt)/(m))) In an experimental set-up four objects I,II,III,IV were released in same liquid. Using the data collected for the subsequent motions value of constant b were calculated. Respective data are shown in table. {:("Object",I,II,II,IV),("Mass (in kg.)",1,2,3,4),(underset("in (N-s)/m")("Constant b"),3.7,1.4,1.4,2.8):} A small sphere of mass 2.00 g is released from rest in a large vessel filled with oil. The sphere approaches a terminal speed of 10.00 cm/s. Time required to achieve speed 6.32 cm/s from start of the motion is (Take g = 10.00 m//s^(2) ) :

When the resistance in left and right gap of a mere bridge are 101Omega and 1Omega respectively, the null point is found at 99.5 cm. When the resistances are interchanged the null point is now at 0.7cm. Calcualte the end-corrections of the metre bridge.

In the meter bridge experiment, the length Ab of the wire is 1m. The resistors X and Y have values 5Omega and 2Omega respectively. When a shunt resistance S is connected to X, the balacing point is found to be 0.625 m from A. Then the resistance of the shunt is