Under special circumstances the Election Commission holds re-elections in a particular constituency for a second time.
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Solution and Explanation
Correct Answer: The Election Commission orders re-elections when irregularities compromise the fairness of the election.
Step 1: Role of the Election Commission in Ensuring Fair Elections
The Election Commission of India (ECI) ensures free, fair, and transparent elections across the country. If irregularities such as booth capturing, rigging, or violence occur, the ECI is empowered to take corrective measures, including ordering re-elections in affected constituencies to protect the sanctity of the process.
Step 2: Reasons for Re-elections
Re-elections are ordered when the fairness of the original voting is compromised due to issues like tampering with Electronic Voting Machines (EVMs), voter intimidation, or disturbances during polling. The ECI can cancel election results and mandate re-polling to uphold democratic integrity.
Step 3: Ensuring the Integrity of Elections
Re-elections reinforce trust in democracy by guaranteeing that voters can cast their votes freely and fairly. They help maintain transparency and legitimacy in the electoral process, ensuring that every citizen’s vote truly counts.
\[ \text{Re-elections are ordered by the ECI to uphold fairness and integrity in the democratic process.} \]
Top Questions on Civics
Questions Asked in Maharashtra Class X Board exam
Complete the following activity to prove that the sum of squares of diagonals of a rhombus is equal to the sum of the squares of the sides.
Given: PQRS is a rhombus. Diagonals PR and SQ intersect each other at point T.
To prove: PS\(^2\) + SR\(^2\) + QR\(^2\) + PQ\(^2\) = PR\(^2\) + QS\(^2\)
Activity: Diagonals of a rhombus bisect each other.
In \(\triangle\)PQS, PT is the median and in \(\triangle\)QRS, RT is the median.
\(\therefore\) by Apollonius theorem,
\[\begin{aligned} PQ^2 + PS^2 &= \boxed{\phantom{X}} + 2QT^2 \quad \dots \text{(I)} \\ QR^2 + SR^2 &= \boxed{\phantom{X}} + 2QT^2 \quad \dots \text{(II)} \\ \text{Adding (I) and (II),} \quad PQ^2 + PS^2 + QR^2 + SR^2 &= 2(PT^2 + \boxed{\phantom{X}}) + 4QT^2 \\ &= 2(PT^2 + \boxed{\phantom{X}}) + 4QT^2 \quad (\text{RT = PT}) \\ &= 4PT^2 + 4QT^2 \\ &= (\boxed{\phantom{X}})^2 + (2QT)^2 \\ \therefore \quad PQ^2 + PS^2 + QR^2 + SR^2 &= PR^2 + \boxed{\phantom{X}} \\ \end{aligned}\]
- Maharashtra Class X Board - 2025
- Quadrilaterals
- Prove that tangent segments drawn from an external point to a circle are congruent.
- Draw a circle with radius 4.1 cm. Construct tangents to the circle from a point at a distance 7.3 cm from the centre.
- Maharashtra Class X Board - 2025
- Constructions
- \(\triangle\)ABC \(\sim\) \(\triangle\)PQR. In \(\triangle\)ABC, AB = 3.6 cm, BC = 4 cm and AC = 4.2 cm. The corresponding sides of \(\triangle\)ABC and \(\triangle\)PQR are in the ratio 2 : 3, construct \(\triangle\)ABC and \(\triangle\)PQR.
- Maharashtra Class X Board - 2025
- Constructions
- \(\square\)ABCD is a rectangle. Taking AD as a diameter, a semicircle AXD is drawn which intersects the diagonal BD at X. If AB = 12 cm, AD = 9 cm, then find the values of BD and BX.