History, asked by StrongGirl, 11 months ago

The Rajya Sabha passed ‘The Mineral Laws (Amendment) Bill, 2020’, which proposes to commercialise the mining of which mineral?
Iron
Bauxite
Gold
Coal

Answers

Answered by PADMINI

0

The Rajya Sabha passed ‘The Mineral Laws (Amendment) Bill, 2020’, which proposes to commercialize the mining of "coal".

Explanation:

Recently, The Rajya Sabha passed The Mineral Laws (Amendment) Bill, 2020.

The Bill proposes amendments to the Mines and Mineral Act 1957 and The Coal Mines Act, 2015.

This bill was passed by Lok Sabha on 6th March 2020. It will open a new era in the Indian coal & mining sector especially to promote Ease of Doing Business.

Answered by Joy369

0

Answer:coal

Explanation:be is a non renewable energy we have used most of it for electricity

Previous Question

Next Question

Similar questions

Hindi, 5 months ago

anopcharik patra for class 5...

Math, 5 months ago

5) If O is the centre of the circle, find the value of x. * 1.) 130° 2.) 210° 3.) 230° 4.) 270°...

Math, 5 months ago

construct a triangle of sides 5cm 6cm and 7cm then a triangle similar to it whose sides 4/5 of the corresponding sides of the first triangle...

English, 11 months ago

name the games which can be played with friends in this long holiday season...

History, 11 months ago

explain the role of nehru in non aligned movement...

Geography, 1 year ago

Draw a parallel between medieval towns and modern day settlements...

Physics, 1 year ago

At what angle must the two forces (x + y) and (x – y) act so that the resultant may be $\sqrt{(x^{2} + y^{2})}$ ? (a) $\cos^{-1}[-(x^{2} + y^{2})}/{2(x^{2} - y^{2})}]$ (b) $\cos^{-1}[-2{(x^{2} - y^{2})}/{(x^{2} + y^{2})}]$ (c) $\cos^{-1}[-{(x^{2} + y^{2})}/{(x^{2} - y^{2})}]$ (d) $\cos^{-1}[-{(x^{2} - y^{2})}/{(x^{2} + y^{2})}]$ ...

Physics, 1 year ago

For any two vectors $\vec{A}\ and\ \vec{B},\ if\ \vec{A}.\vec{B}=|\vec{A}\ \times\ \vec{B}|$ , the magnitude of $\vec{C} = \vec{A} + \vec{B}$ is (a) A + B (b) $\sqrt{A^{2}+B^{2}+\sqrt{2}AB}$ (c) $\sqrt{A^{2}+B^{2}}$ (d) v $\sqrt{A^{2}+B^{2}+\frac{AB}{\sqrt{2}}}$ ...