Bank Exams Questions

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Intermolecular forces :

The forces of attraction present in between the molecules or atoms or compounds is termed as intermolecular forces.


The molecule HCl is composed of hydrogen and chlorine atom. There is high difference in electronegativities therefore, it is polar in nature. The H end of HCl is permanently slightly positive charge. The Cl end of HCl has a permanent slight negative charge. the "H" in one molecule is attracted to the "Cl" in a neighbor. The intermolecular force is weak compared to a covalent bond. But this dipole-dipole interaction is one of the stronger intermolecular attractions.

It also has hydrogen bonding and dispersion forces. 

Dispersion forces are the weakest forces which are present in between all molecules or atoms.
Hydrogen bonds are present in molecules having at least one H-A bond where A is the strong electronegative atom (mainly F, Cl and N).

A variable provides us with named storage that our programs can manipulate. Each variable in C++ has a specific type, which determines the size and layout of the variable's memory the range of values that can be stored within that memory and the set of operations that can be applied to the variable.

The name of a variable can be composed of letters, digits, and the underscore character. It must begin with either a letter or an underscore. Upper and lowercase letters are distinct because C++ is case-sensitive −

Basic types of variables:

1. bool

Stores either value true or false.

2. char

Typically a single octet (one byte). This is an integer type.

3. int

The most natural size of an integer for the machine.

4. float

A single-precision floating point value.

5. double

A double-precision floating point value.

6. void

Represents the absence of type.

7. wchar_t

A wide character type.

The distance is constant in this case.

Let the time taken for travel with a speed of 10 kmph be 't'.

Now the speed of 15 kmph is 3/2 times the speed of 10 kmph.

Therefore, time taken with the speed of 15 kmph will be 2t/3 (speed is inversely proportional to time)

Extra time taken = t - 2t/3 = t/3

=> 1pm - 11am = 2hrs

=> t/3 = 2h

=> t = 6 hrs.

Now, Distance = speed x time = 10 x 6 = 60 kms

Time he takes to reach at noon = 6 - 1 = 5 hrs

Now, Speed = 60/5 = 12 kmph.

Given speed of jogger = 9 kmph
Speed of train = 45 kmph
As they are in same direction,
Relative speed = 45 - 9 = 36 kmph = 36x5/18 = 10 m/s
Distance = 240 + 120 = 360 mts
Time = distance/speed = 360/10 = 36 sec.