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Given problem is a typical mixing problem

dy dt ratein rategut

Let y(t) is the salt in tank at time t

The water entering has no salt, rate in =0

The thoroughly mixed solution has concentration of 10L/min . The tank has 1119 L of brine with 15 kg of salt

Rate out =( y/1119) kg/ L * 10 L/min = 10 y/ 1119 kg/mim

Hence, differential equation is dy dt 10y 1119

dy Y 10 -dt 1119

dy =-11 10 •dt 1119 S

10 In y = - + tc 1119

y = e 1119 +c

at t=0, y= 15

15= C

c= 15

Therefore, y= 10 y = 15e 1119 =

After 20 minutes, t= 20

10.200 y = 15e 1119 kg

y= 12.54496 kg

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QUESTION 6 (a) The bar shown in Figure Q2(a) is subjected to tensile load of 150 Kn. If the stress in the middle portions is limited to 160 N/mm², determine the diameter of the middle portion. Find also the length of the middle portion if the total elongation of the bar is to be 0.25 mm. E E = 2.0 + 105N/mm². (12 marks) 150 KN 10 cm DIA 10 cm DIA 150 KN 45 cm Figure Q6(a) (b) A brass bar, having cross-section area of 900 mm², is subjected to axial forces as shown in Figure Q2(b), in which AB = 0.6 m, BC = 0.8 m, and CD = 1.0 m. Find the total elongation of the bar. E = 1.0 + 105N/mm2 . (8 marks) Page 4 of 5 B D 40 KN 70 KN 20 KN 10 KN Figure Q6(b) (TOTAL = 20 MARKS)

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A palindrome is a number or a text phrase that reads the same backwards as forwards. For example, each of the following five-digit integers is a palindrome: 12321, 55555, 45554 and 11611. Write a C++ program that reads in a five-digit integer and determines whether it is a palindrome .

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Question: A palindrome is a number or a text phrase that reads the same backwards as forwards. For example, each of the following five-digit integers is a palindrome: 12321, 55555, 45554 and 11611. Write a C++ program that reads in a five-digit integer and determines whether it is a palindrome . Answer: Step 1 PROGRAMING CODE:   #include <stdio.h> int main() {   int n, a, b, c, d, e;   int digit = 0;   printf( "Enter one 5 digit number: \n" );   scanf( "%d", &n );   a = n / 10000;   b = n / 1000 % 10;   c = n / 100 % 10;   d = n / 10 % 10;   e = n % 10;   if ( a == e && b == d ) {     printf( "Palindrome!\n" );   }   if ( a != e || b != d ) {     printf( "Not a palindrome.. :(\n" );   }   return 0; } PROGRAM SCREENSHOT: Step 2
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