**Answer:**

**158.3 kg/s**

**Explanation:**

given,

time to drain water from container = 10 h

volume of water in the container, V = 5700 m³

density of water = 1000 Kg/m³

mass flow rate = ?

mass of water = ρ V

m = 1000 x 5700

m = 5.7 x 10⁶ kg

time of flow = 10 h

= 10 x 60 x 60 = 36000 s

the mass flow rate =

=

= **158.3 kg/s**

hence,** the mass flow rate is equal to 158.3 Kg/s**

<u>**Answer:**</u>

**Option(b)**

The position of the object placed so that the image formed is a real image of the same size as the object is at a distance of **twice** the **focal length**

<u>**Explanation:**</u>

In this case, if the object is to be placed in such a way that the image thus formed by the object is real image of the same size so, the position of the object is at a distance twice of the focal length .

So the object is said to be in position at a distance of **two times** of its focal length i.e., it can be represented as **2f **

Answer:

13.51 nm

Explanation:

To solve this problem, we are going to use angle approximation that sin θ ≈ tan θ ≈ θ where our θ is in radians

y/L=tan θ ≈ θ

and ∆θ ≈∆y/L

Where ∆y= wavelength distance= 2.92 mm =0.00292m

L=screen distance= 2.40 m

=0.00292m/2.40m

=0.001217 rad

The grating spacing is d = (90000 lines/m)^−1

=1.11 × 10−5 m.

the small-angle

approx. Using difraction formula with m = 1 gives:

mλ = d sin θ ≈ dθ →

∆λ ≈ d∆θ = =1.11 × 10^-5 m×0.001217 rad

=0.000000001351m

= 13.51 nm

**Answer:**

The resistance of the heater wire is of R= 0.68 Ω.

**Explanation:**

1 cal/s = 4.184 W

P= 50 cal/s = 209.2 W

V= 12V

P= V* I

I= P/V

I= 17.43 A

P= I² * R

R= P / I²

**R= 0.68 Ω**