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The relation between T and g is given by
A.TgB.Tg2C.T2gD.Tlg

Answer
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Hint: In the question we are asked to find the relation between ‘T’ and ‘g’. Consider a mass “m” suspended on a wire of length ‘l ’ undergoes simple harmonic motion, ‘T’ is the time period, the time required to complete one oscillation and ‘g’ is the acceleration due to gravity (9.8m/s). To solve this we know the equation of time period; by squaring the known equation we can formulate the relation between ‘T’ and ‘g’.

Formula used:
T=2πlg
Time period of a simple oscillation is given by

Complete answer:
To find the relation between ‘T’ and ‘g’,
Let us consider ‘l ’ to be the length of the pendulum.
As we know, time period is given by the equation
T=2πlg
Squaring both sides of the equation, we get
T2=4π2lg
From this equation we get,
T2lg
Thus we can conclude that, Tlg, when l is unchanged.

So, the correct answer is “Option D”.

Additional Information:
Time of a simple pendulum derivation:

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Consider a simple pendulum with a mass ‘m’ suspended on a wire of length ‘l ’.
For one oscillation the pendulum is displaced at an angle of ‘θ’ by ‘x’ distance.
Let T0 be the time period at equilibrium.
T0=mg
When the pendulum oscillates, it is displaced at a small angle θ
For this small displacement θ, the restoring force acting will be
Restoring force=mgsinθ
Since the angle of displacement θ is very small here, we can approximate sinθ to θ
I.e. sinθθ
Hence the force here can be rewritten as
F=mgθ
Now let us consider the triangle ABC in the figure.
We know that sin of the angle θ is the ratio of the opposite side to the hypotenuse of the triangle. Since here sinθθ, we can write this as
θ=oppositehypotenuse
Here the opposite side of the angle is the displacement ‘x’ and the hypotenuse of the triangle is length ‘l ’ of the pendulum. Hence we can rewrite the equation as
θ=xl
Therefore the restoring force on the pendulum is
F=mgθ=mg×xl
By Newton’s second law of motion, we have the equation of motion as
F=ma, Where ‘m’ is the mass of the body and ‘a’ is the acceleration.
We can rewrite this equation as
a=Fm
From the previous equation, we know that F=mgxl. Substituting this here, we get
a=mg(xl)m
Eliminating the common terms, we get
a=gl×x
For a simple harmonic motion we know that, a=ω2x
On comparing both these equations, we get
ω2x=glx
By simplifying this,
ω2=gl
ω=gl
Time period ‘T’ is given by the equation
T=2πω
Substitute the value of ω in this equation
Therefore time period, T=2πlg

Note:
This question can be solved by another method.
We know that, for a simple pendulum its angular frequency ω is given by
ω=gl
Time period of an oscillation can also be written as
T=2πω
By substituting the value of angular frequency (ω ) in the above equation, we get
T=2πlg
Thus we get Tlg
Hence we get the same solution.