State the norms that are set by the Election Commission to get recognition as a national party
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Solution and Explanation
Correct Answer: Norms set by the Election Commission to get recognition as a national party.
Step 1: Definition of a National Party
A political party in India is recognized as a national party if it meets the Election Commission of India (ECI) criteria that demonstrate broad, multi-state support. National recognition brings benefits like a common symbol across India, priority access to electoral rolls/symbols, and greater visibility/resources.
Step 2: Criteria for Recognition (any one of the following)
- Vote Share Path: Secures ≥ 6% of valid votes in at least four states in Lok Sabha or State Assembly elections and wins at least 4 Lok Sabha seats.
- Seat Share Path: Wins ≥ 2% of Lok Sabha seats (i.e., at least 11 seats) from at least three states.
- Multi-State Recognition Path: Recognized as a state party in at least four states.
Step 3: Importance of National Party Status
National status enables a party to contest nationwide with a single symbol, improves administrative access, and boosts credibility for alliances, fundraising, and voter outreach, thereby enhancing its national footprint.
\[ \textbf{Summary: }\; \text{Meet votes + seats thresholds, or hold state-party status in } \ge 4 \text{ states to qualify as a national party.} \]
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.