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# Tutor profile: Satyam D.

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Satyam D.
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## Questions

### Subject:Physics (Electricity and Magnetism)

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Question:

A bar magnet of magnetic moment $$2.0 Am^{2}$$ is free to rotate about a vertical axis through its centre. The magnet is released from rest from the east-west position. Find the kinetic energy of the magnet as it takes the north-south position. The horizontal component of the Earth’s magnetic field is, B = 25 µT.

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Satyam D.

Given: A bar magnet placed in a magnetic field perpendicular to its alignment is released from rest. Magnetic moment of bar magnet, M = 2.0 Am2 Earth’s magnetic field, B = 25 µT To find: Kinetic energy of the bar magnet when it is aligned with the magnetic field. Concept: A bar magnet (or, a magnetic dipole) placed in a uniform magnetic field, experiences a torque. The magnetic field exerts a torque which tends to align the dipole in its direction. In this case, the dipole is perpendicular to the magnetic field. The potential energy, U, of a dipole of magnetic moment, M, in a magnetic field, B, placed at an angle Ө with the magnetic field, is given by $$U = -MBcosӨ$$ The bar magnet rotates to decrease its potential energy and the decrease in potential energy leads to increase in kinetic energy. We need to find the kinetic energy when the dipole is aligned with the magnetic field. Solution: $$U$$ = Potential energy of the dipole $$K$$ = Kinetic energy of the dipole Initially, $$U_{i} = -MBcosӨ = -(2)(25)(cos(π/2)) = 0$$ $$K_{i} = 0$$ Finally, $$U_{f} = -MBcosӨ = -(2)(25)(cos(0)) = -50 µJ$$ $$K_{f} = ?$$ Since the total mechanical energy of the dipole remains conserved, therefore, $$U_{i} + K_{i} = U_{f} + K_{f}$$ $$0 + 0 = -50 + K_{f}$$ $$K_{f} = 50 µJ$$ Hence, the kinetic energy of the bar magnet when it is aligned with the magnetic field is 50 µJ.

### Subject:Physical Chemistry

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Question:

The following data were obtained during the first order thermal decomposition of $$N_2O_5$$(g) at constant volume: $$N_2O_5\rightarrow2N_2O_4 + O_2$$ S.NO. time (s) total pressure(atm) 1. 0 0.5 2. 100 0.512 Calculate the rate constant.

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Satyam D.

Given: Balanced chemical equation of decomposition reaction of $$N_2O_5$$. At time t=0, Total pressure, $$P_{i} = 0.5 atm$$ At time t=100, Total pressure, $$P_{f}=0.512 atm$$ To find: Rate constant of the reaction $$k$$ Concepts: The given reaction is a first order reaction. A first order reaction depends on the concentration of only one reactant (a unimolecular reaction). Other reactants can be present, but each will be zero order. For a first order reaction, $$A \rightarrow B$$ if concentration of A = [A], rate of reaction = r, rate constant of reaction = k the rate law is given by $$-\frac{d[A]}{dt} = r = k[A]$$ k is the first order rate constant, which has units of $$\frac{1}{s}$$ or $$s^{−1}$$. The integrated first order rate law is $$log[A]_{t} = -kT + log[A]_{0}$$ $$where [A]_{t}$$ = concentration of A at time t, $$[A]_{0}$$= concentration of A at t=0 $$\Rightarrow k=\frac{2.303}{t}log\frac{[A]_{0}}{[A]_{t}}$$ For gaseous reactions in closed containers(constant volume), the pressure of the reactants and products is directly proportional to the corresponding no. of moles of of the reactants and products. $$\Rightarrow k=\frac{2.303}{t}log\frac{p_i}{p_A}$$ where $$p_{i}$$ = initial partial pressure of A, $$p_{A}$$ = partial pressure of A at time = t Solution: The pressure of $$N_2O_5$$ will decrease by 2x atm and pressure of $$N_2O_4$$ will increase by 2x atm and pressure of $$O_2$$ will increase by x atm , as two moles of $$N_2O_5$$ gives 2 moles of $$N_2O_4$$ and 1 mole of $$O_2$$. $$k=\frac{2.303}{t}log\frac{p_i}{p_A}$$ $${2N_{2}O_{5}}_{(g)} \rightarrow {2N_{2}O_4}_{(g)} + {O_2}_{(g)}$$ start t=0 0.5 atm 0 atm 0 atm time t=100 (0.5-2x) atm 2x atm x atm $$P_{f}=p_{N_2O_5} + p_{N_2O_4} + p_{O_2}$$ $$=(0.5-2x) + 2x + x$$ $$=0.5 + x$$ $$x=P_{f}-0.5$$ $$x=0.012$$ $$p_{N_2O_5}=0.5-2x$$ $$p_{N_2O_5}=0.476$$ $$k=\frac{2.303}{t}log\frac{p_i}{p_A}$$ $$k=\frac{2.303}{100}log\frac{0.5}{0.476}$$ $$k=\frac{2.303}{100} \times 0.0216$$ $$k=4.98 \times 10^{-4} s^{-1}$$ Thus, the rate constant of the reaction is $$4.98 \times 10^{-4} s^{-1}$$

### Subject:C Programming

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Question:

Write a C program which takes a string as input from the user and outputs the no. of upper case letters, lower case letters and special characters in the input string.

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Satyam D.

In order to write a working and clean code, it is a good practice to divide the problem into smaller problems which can individually be solved to get to the solution. We start by writing the steps, as can be seen by looking at the problem. We then keep dividing the steps into smaller steps that can be simply translated to simple, individual blocks of code. By sequentially arranging these individual blocks of code after checking for boundary cases and exceptions (if any), in most of the cases, we get the required code for the problem statement. Procedure: - Get a string of characters as input from the user - Parse the string character-wise - For each character, check if it is upper case, lower case or a special character - Increase the count of uppercase, lowercase, or special characters , depending on what the character is - Loop through all the characters of the string and repeat the process - Print the final counts of uppercase, lowercase and special characters After writing the steps, it’s easy to write the inputs and outputs required by the program. Input: A string of characters Output: For the input string, print 1. No. of uppercase letters 2. No. of lowercase letters 3. No. of special characters Boundary cases/exceptions: None Required variables: - User input: array of characters - Uppercase count: integer - Lowercase count: integer - Special characters count: integer - Loop counter: integer Code: #include<stdio.h> int main() { // declare variables char text; //input string int i; // loop counter int countL,countU,countS; // ask for string input printf("Enter any string: "); gets(text); //here, we are printing string using printf //without using loop printf("Entered string is: %s\n",text); //count lower case, upper case and special characters //assign 0 to counter variables countL=countU=countS=0; for(i=0;text[i]!='\0';i++) { //check for alphabet if ((text[i]>='A' && text[i]<='Z') || (text[i]>='a' && text[i]<='z')) { if ((text[i]>='A') && (text[i]<='Z')) { //it is upper case alphabet countU++; } else { //it is lower case character countL++; } } else { //character is not an alphabet countS++; //it is special character } } //print values printf("Upper case characters: %d\n",countU); printf("Lower case characters: %d\n",countL); printf("Special characters: %d\n",countS); return 0; }

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