Question

$\bf{\fbox\red {PHYSICS}}$

⤥⟨ Explain your answer ⟩♣


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Answer 1

SOLUTION

Cost of oranges and mangoes, each per kg=Rs.40

Amount paid daily for mangoes and oranges each =Rs.1200

⇒120040=30kg

Thus, the merchant got 30kg of oranges and mangoes each that day.

Let the price of mango be Rs.x next day.

⇒ Price of orange next day = Rs.(x+10

From the information given in the problem, we have

1200x−1200x+10=20

⇒60x−60x+10=1

60x+600−60x=x(x−10)

⇒600=x2+10x

⇒x2+10x−600=0

⇒x2+30x−20x−600=0

⇒x(x+30)−20(x+30)=0

⇒(x+30)(x−20)=0

⇒x=−30,20

−30 is inadmissible , hence x=20

⇒x+10=20+10=30

Thus, price of oranges =Rs.30/kg on that day.

Answer 2

Answer:

(4) 2/√5 s is the correct answer

Explanation:

For physics most of the questions are formula based. So, we just need to remember the terms present in the formula

Given :

Area of container = 10 × Area of the hole

A = 10a

To find :

The time taken by water level to get lowered from the height of 81 cm to 49 cm

Formula :

$\Large \text {{ t = \frac{A}{a}\sqrt{\frac{2}{g}} \ (\sqrt{H_1} - \sqrt{H_2} ) }}$

t = time taken for the water to get lowered from the height of H₁ to H₂

A = Area of the container

a = Area of the hole

g = acceleration due to gravity = 10 m/s²

H₁ and H₂ = the heights from where the water starts flowing out(Need to measured from the ground)

[Here, we may get confused of the height whether we must take the height from the top of the container or from the ground. So, the point should be noted that, we must consider the Height from the ground]

Solution :

So, substituting the available terms in the formula,

$t = \frac{10a}{a} \sqrt{\frac{2}{10} } \ (\sqrt{81 \times 10^{-2}} - \sqrt{49 \times 10^{-2}} )$

In the question it is mentioned as cm, which we must convert into m

So, 81 cm = 81 × 10 ⁻² m and 49 cm = 49 × 10⁻² m

$t = 10 \sqrt{\frac{1}{5} } \ [( 9 - 7) \times 10^{-1}]\\\\ \boxed {t = \frac{2}{\sqrt{5} } s}$

Hope it helps!