Algorithm

1. Right shift by n times to get the nth bit at LSB
2. Do a bitwise and with 1(only LSB is set of 1, other bits 0).
3. IF result is 1, then nth bit is set
   Else
   Bit not set

Example with explanation:

Let input no be 67600 and n is 11

This is the binary representation of 67600

After right shifting 11 times,

So bitwise ANDing this shifted no with 1

Results in 00000...0001(31 0's & one 1) which is 1... Thus n th bit is set...

C implementation to Check if nth Bit in a 32-bit Integer is set or not

#include <stdio.h>

int main()
{   
    int n,k;
    printf("enter a 32 bit number\n");
    scanf("%d",&k);
    printf("enter the bit no to check...\n");
    printf("bit-no 0-indexed & 0 starts from LSB...\n");
    scanf("%d",&n);
	if(n>32){
		printf("enter between 0-31\n");
		return;
	}
    
    k=k>>n; //right shift the no to get nth bit at LSB
    if(k&1==1) //check whether nth bit set or not
    printf("%d th bit is set\n",n);
    else
    printf("%d th bit not set\n",n);
    
    return 0;
    
}

Output

First run:
enter a 32 bit number
67600
enter the bit no to check...
bitno 0-indexed & 0 starts from LSB...
11
11 th bit is set


Second run:
enter a 32 bit number
67600
enter the bit no to check...
bit-no 0-indexed & 0 starts from LSB...
10
10 th bit not set