Numerical Methods - Engineering Mathematics - Instrumentation Engineering Previous Year Questions

1

2009 · Instrumentation Engineering · Engineering Mathematics · Numerical Methods

Instrumentation Engineering (IN) 2009

The differential equation \frac{dx}{dt} = \frac{4 - x}{\tau}, with x(0) = 0, and the constant τ > 0, is to be numerically integrated using the forward Euler method with a constant integration time step T. The maximum value of T such that the numerical solution of x converges is

A

τ / 4

B

τ / 2

C

τ

D

2τ

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2

2010 · Instrumentation Engineering · Engineering Mathematics · Numerical Methods

Instrumentation Engineering (IN) 2010

The velocity \(v\) (in m/s) of a moving mass, starting from rest, is given as \(\frac{dv}{dt} = v + t\). Using Euler forward difference method (also known as Cauchy-Euler method) with a step size of 0.1 s, the velocity at 0.2 s evaluates to

A

0.01 m/s

B

0.1 m/s

C

0.2 m/s

D

1 m/s

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3

2010 · Instrumentation Engineering · Engineering Mathematics · Numerical Methods

Instrumentation Engineering (IN) 2010

5 skilled workers can build a wall in 20 days; 8 semi-skilled workers can build a wall in 25 days; 10 unskilled workers can build a wall in 30 days. If a team has 2 skilled, 6 semi-skilled and 5 unskilled workers, how long will it take to build the wall?

A

20 days

B

18 days

C

16 days

D

15 days

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4

2011 · Instrumentation Engineering · Engineering Mathematics · Numerical Methods

Instrumentation Engineering (IN) 2011

The extremum (minimum or maximum) point of a function \(f(x)\) is to be determined by solving \(\frac{d f(x)}{d x} = 0\) using the Newton-Raphson method. Let \(f(x) = x^3 - 6x\) and \(x_0 = 1\) be the initial guess of \(x\). The value of \(x\) after two iterations (\(x_2\)) is

A

0.0141

B

1.4142

C

1.4167

D

1.5000

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5

2013 · Instrumentation Engineering · Engineering Mathematics · Numerical Methods

Instrumentation Engineering (IN) 2013 [Session 1]

While numerically solving the differential equation \(\frac{dy}{dx} + 2xy^2 = 0\), \(y(0) = 1\) using Euler’s predictor-corrector (improved Euler-Cauchy) method with a step size of 0.2, the value of \(y\) after the first step is

A

1.00

B

1.03

C

0.97

D

0.96

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6

2013 · Instrumentation Engineering · Engineering Mathematics · Numerical Methods

Instrumentation Engineering (IN) 2013 [Session 2]

While numerically solving the differential equation \( \frac{dy}{dx} + 2xy^2 = 0 \), \( y(0) = 1 \) using Euler's predictor-corrector (improved Euler-Cauchy) method with a step size of 0.2, the value of y after the first step is

A

1.00

B

1.03

C

0.97

D

0.96

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7

2013 · Instrumentation Engineering · Engineering Mathematics · Numerical Methods

Instrumentation Engineering (IN) 2013 [Session 3]

While numerically solving the differential equation \( \frac{dy}{dx} + 2xy^2 = 0 \), y(0)=1 using Euler's predictor-corrector (improved Euler-Cauchy) method with a step size of 0.2, the value of y after the first step is

A

1.00

B

1.03

C

0.97

D

0.96

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8

2013 · Instrumentation Engineering · Engineering Mathematics · Numerical Methods

Instrumentation Engineering (IN) 2013 [Session 4]

While numerically solving the differential equation \(\frac{dy}{dx} + 2xy^2 = 0\), \(y(0)=1\) using Euler's predictor-corrector (improved Euler-Cauchy) method with a step size of 0.2, the value of \(y\) after the first step is

A

1.00

B

1.03

C

0.97

D

0.96

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9

2014 · Instrumentation Engineering · Engineering Mathematics · Numerical Methods

Instrumentation Engineering (IN) 2014

The iteration step in order to solve for the cube roots of a given number \( N \) using the Newton-Raphson's method is

A

\(x_{k+1} = x_k + \frac{1}{3} (N - x_k^3) \)

B

\(x_{k+1} = \frac{1}{3} \left( 2x_k + \frac{N}{x_k^2} \right) \)

C

\(x_{k+1} = x_k - \frac{1}{3} (N - x_k^3) \)

D

\(x_{k+1} = \frac{1}{3} \left( 2x_k - \frac{N}{x_k^2} \right) \)

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10

2024 · Instrumentation Engineering · Engineering Mathematics · Numerical Methods

Instrumentation Engineering (IN) 2024

Consider a system given by the following first order differential equation:
\(\frac{dy}{dt} = y + 2t - t^2\)
where, y(0) = 1 and 0 ≤ t < ∞. Using a step size h = 0.1 for the improved Euler method, the value of y(t) at t = 0.1 is ______ (rounded off to two decimal places).

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11

2025 · Instrumentation Engineering · Engineering Mathematics · Numerical Methods

Instrumentation Engineering (IN) 2025

Newton-Raphson method is used to compute the inverse of the number 1.6. Among the following options, the initial guess of the solution that results in non-convergence of the iterative process is

A

0.55

B

0.75

C

1.15

D

1.25

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12

2007 · Instrumentation Engineering · Engineering Mathematics · Numerical Methods

Instrumentation Engineering (IN) 2007

Identify the Newton-Raphson iteration scheme for finding the square root of 2.

A

\(x_{n+1} = \frac{1}{2} \left( x_n + \frac{2}{x_n} \right)\)

B

\(x_{n+1} = \frac{1}{2} \left( x_n - \frac{2}{x_n} \right)\)

C

\(x_{n+1} = \frac{2}{x_n}\)

D

\(x_{n+1} = \sqrt{2 + x_n}\)

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13

2017 · Instrumentation Engineering · Engineering Mathematics · Numerical Methods

Instrumentation Engineering (IN) 2017

The following table lists an \(n^{th}\) order polynomial \(f(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0\) and the forward differences evaluated at equally spaced values of \(x\). The order of the polynomial is

A

1

B

2

C

3

D

4

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14

2020 · Instrumentation Engineering · Engineering Mathematics · Numerical Methods

Instrumentation Engineering (IN) 2020

Consider the recursive equation \( X_{n+1} = X_n - h(F(X_n) - X_n) \), with initial condition \( X_0 = 1 \) and \( h > 0 \) being a very small valued scalar. This recursion numerically solves the ordinary differential equation ____.

A

\(\dot{X} = -F(X), X(0) = 1 \)

B

\(\dot{X} = -F(X) + X, X(0) = 1 \)

C

\(\dot{X} = F(X), X(0) = 1 \)

D

\(\dot{X} = F(X) + X, X(0) = 1 \)

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15

2022 · Instrumentation Engineering · Engineering Mathematics · Numerical Methods

Instrumentation Engineering (IN) 2022

The Newton-Raphson method is applied to determine the solution of f(x) = 0 where f(x) = x - cos(x). If the initial guess of the solution is x₀ = 0, the value of the next approximation x₁ is ______ (round off to two decimal places)

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