Question

Two poles of heights 16 m and 7m stand on a plane ground . If the distance between their feet is 12m , the distance between their tops is​

Answer 1

Answer:

Distance between their tops is 15 metres

Explanation:

Given, two poles of 16m and 7m standing on a plane ground. The distance between their feet is 12m.

We need to find the distance between their tops.

Following the figure,

We have,

• AB and DC are the two poles of height 7m and 16m respectively.
• AD is the distance between their feet = 12m
• BC denotes the distance between their tops.

And, we need to find BC

In ∆BCE,

• BE = AD = 12m
• EC = DC - AB = 16 - 7 = 9m

Applying Pythagoras theorem,

⇒ (BC)² = (BE) + (EC)

⇒ (x)² = (12)² + (9)²

⇒ x² = 144 + 81

⇒ x² = 225

⇒ x = √225

⇒ x = 15

∴ Distance between their tops is 15m

Answer 2

Answer:

⇝ We Know :-

Mass of Magnesium (Mg) = 24 u

Mass of Sulphur (S) = 32 u

Mass of Oxygen (O) = 16 u

Mass of Hydrogen (H) = 1 u

⇝ Calculating Molar Mass of MgSO₄.7H₂O :

Molar Mass = Mass of Mg + Mass of Sulphur + 11 × Mass of Oxygen + 14 × Mass of Hydrogen.

❒ Using Above written Masses :

\begin{gathered}\text{ Molar mass} = 24 + 32 + 11 \times 16 + 14 \times 1 \\ \end{gathered} Molar mass=24+32+11×16+14×1

\begin{gathered} = 24 + 32 + 176 + 14 \\ \end{gathered}=24+32+176+14

\large\red{ \bf = 246 \: u}=246u

⇝ Calculating Molar Mass of 7H₂O :

Molar Mass = 7 × ( 2× Mass of Hydrogen + Mass of Oxygen)

❒ Using Respective Masses :

\begin{gathered}\text{ Molar mass} = 7 \times(2 \times 1 +16)\\ \end{gathered} Molar mass=7×(2×1+16)

\begin{gathered} =7\times( 18) \\ \end{gathered}=7×(18)

\large\red{ \bf = 126 \: u}=126u

⇝ Calculating Reqrd. percentage :-

\begin{gathered} \purple{\sf \small \% \: of \: water = \frac{mass \: of \: water \: in \: molecule}{mass \: of \: molecule \: } \times 100} \\ \end{gathered}%ofwater=massofmoleculemassofwaterinmolecule×100

\begin{gathered} = \frac{126}{246} \times 100 \\ \end{gathered}=246126×100

\begin{gathered} = 0.5121 \times 100 \\ \end{gathered}=0.5121×100

\bf= 51.21\%=51.21%

Hence,

\large\underline{\pink{\underline{\frak{\pmb{Required \: \% = 51.21\%}}}}}Required%=51.21%Required%=51.21%

Step-by-step explanation:

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