1. The void-pressure diagram is shown above. What is the coefficient of compressibility ?





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MCQ->What will be the output of the following program? #include<iostream.h> class Bix { int x, y; public: void show(void); void main(void); }; void Bix::show(void) { Bix b; b.x = 2; b.y = 4; cout<< x << " " << y; } void Bix::main(void) { Bix b; b.x = 6; b.y = 8; b.show(); } int main(int argc, char argv[]) { Bix run; run.main(); return 0; }....
MCQ->What will be the output of the following program? #include<iostream.h> struct IndiaBix { int arr[5]; public: void BixFunction(void); void Display(void); }; void IndiaBix::Display(void) { for(int i = 0; i < 5; i++) cout<< arr[i] << " " ; } void IndiaBix::BixFunction(void) { static int i = 0, j = 4; int tmp = arr[i]; arr[i] = arr[j]; arr[j] = tmp ; i++; j--; if(j != i) BixFunction(); } int main() { IndiaBix objBix = {{ 5, 6, 3, 9, 0 }}; objBix.BixFunction(); objBix.Display(); return 0; }....
MCQ->The void-pressure diagram is shown above. What is the coefficient of compressibility ?....
MCQ-> Applicants for the doctoral programmes of Ambi Institute of Engineering (AIE) and Bambi Institute of Engineering (BIE) have to appear for a Common Entrance Test (CET). The test has three sections: Physics (P), Chemistry (C), and Maths (M). Among those appearing for CET, those at or above the 80th percentile in at least two sections, and at or above the 90th percentile overall, are selected for Advanced Entrance Test (AET) conducted by AIE. AET is used by AIE for final selection.For the 200 candidates who are at or above the 90th percentile overall based on CET, the following are known about their performance in CET: 1. No one is below the 80th percentile in all 3 sections. 2. 150 are at or above the 80th percentile in exactly two sections. 3. The number of candidates at or above the 80th percentile only in P is the same as the number of candidates at or above the 80th percentile only in C. The same is the number of candidates at or above the 80th percentile only in M. 4. Number of candidates below 80th percentile in P: Number of candidates below 80th percentile in C: Number of candidates below 80th percentile in M = 4:2:1.BIE uses a different process for selection. If any candidate is appearing in the AET by AIE, BIE considers their AET score for final selection provided the candidate is at or above the 80th percentile in P. Any other candidate at or above the 80th percentile in P in CET, but who is not eligible for the AET, is required to appear in a separate test to be conducted by BIE for being considered for final selection. Altogether, there are 400 candidates this year who are at or above the 80th percentile in P.What best can be concluded about the number of candidates sitting for the separate test for BIE who were at or above the 90th percentile overall in CET?
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MCQ->What will be the output of the program? #include<stdio.h> #include<stdarg.h> void fun1(char, int, int , float , char ); void fun2(char ch, ...); void (p1)(char, int, int , float , char ); void (p2)(char ch, ...); int main() { char ch='A'; int i=10; float f=3.14; char p="Hello"; p1=fun1; p2=fun2; (p1)(ch, i, &i, &f, p); (p2)(ch, i, &i, &f, p); return 0; } void fun1(char ch, int i, int pi, float pf, char p) { printf("%c %d %d %f %s \n", ch, i, pi, pf, p); } void fun2(char ch, ...) { int i, pi; float pf; char p; va_list list; printf("%c ", ch); va_start(list, ch); i = va_arg(list, int); printf("%d ", i); pi = va_arg(list, int); printf("%d ", pi); pf = va_arg(list, float); printf("%f ", pf); p = va_arg(list, char ); printf("%s", p); }....
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