Crack the Problem

(25)

Submission: Closed

Solution (.pdf)

Cracked by:

1. Umang Merwana, Bhopal (M.P.).
2. Yash Agrawal, Bhopal (M.P.).

(24)

Submission: Closed

Cracked by:

1. TusharJain, Babu Banarsidas Institute of Technology, Gaziabad, U.P.
2. Vipul Agarwal, Class XII'th, DPS, Gaziabad, U.P.

Vipul's Solution Edited by INDIANET (.pdf)

(23) What is the IUPAC name of the compound:

Submission: Closed

This problem needed application of "First point of difference" rule. For details about the same refer INDIANET's lesson on nomenclature. Solution is provided by : Gaurav Mathur, XII'th, Maheshwari Sr. Secondary School, Jodhpur. Click here for the solution

(22)

Submission: Closed

Cracked by:

Nachiket Vasant Vaidya, 12th, Sathaye College, Vile Parle, Mumbai.

Solution Provided by Nachiket:

Let the spring compress a distance x due to the given force F. So By Work-energy theorem we have,

\(\frac {1}{2}Mass\times Velocity^2 =Fx-\frac {1}{2} Kx^2+(\frac{1}{2}C_2V^2-\frac {1}{2} C_1V^2)\) ............(1)
Where C₁ =4ea²/d and C₂=xae/d +4e(a-x)a/d

We know that dielectric experiences a force inwards given by 1\2V²ea(k-1)/d.(k=dielectric constant) ( This can be easily got by -dU=Force*dx) when F tries to displace it .

So, in the equilibrium of forces, we have displacement s
F=Ks+ 1\2V² (3ea/d) (V=Voltage).... ..............................(2)

So, we get s i.e. equilibrium position where no net force act on it but from (1) it possesses some K.E., i.e. velocity i.e. momentum. But as the other two forces remain constant in magnitude the unbalanced spring force tries to bring the dielectric in the equilibrium position such that F(net )= -Kx where x is the displacement from the Mean Position . Hence it performs SHM.

The Amplitude can be got from (1)

\(0=FA-\frac {1}{2} KA^2+\frac {1}{2}V^2 \frac {-3\epsilon_0aA}{d}\)

So, on Solving we get
A=2F/K-(3eaV²)/Kd

Again From (2) we have s= F/K - 3eaAV²/2Kd

So, the amplitude is A-s i.e.
A/2 = F/K - 3eaAV²/2Kd
And Time Period is given by \(2\pi\sqrt {\frac {M}{K}}\)

(21)

Submission: Closed

Answer: \(By\sqrt {\frac {8\omega}{a}}\)

Cracked by:

1. Yash Agrawal, Bhopal (M.P.).
2. Umang Merwana, Bhopal (M.P.).
3. TusharJain, Babu Banarsidas Institute of Technology Gaziabad, U.P.
4. Jathin Das, Class XII, Abu Dhabi Indian School, Abu Dhabi, UAE.
5. Gaurav Rajput, Class XII, Campion School, Bhopal (M.P.).